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NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Three Dimensional Geometry

Question 1: If a line makes angles 90°, 135°, 45° with xy and z-axes respectively, find its direction cosines.

ANSWER : - Let direction cosines of the line be lm, and n.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the direction Cosines of the line are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 2: Find the direction cosines of a line which makes equal angles with the coordinate axes.

ANSWER : - Let the direction cosines of the line make an angle α with each of the coordinate axes

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, direction cosines of the line make an angle α with each of the coordinate axesNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 3: If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

ANSWER : - If a line has direction ratios of −18, 12, and −4, then its direction cosines are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the direction cosines are -9/11, 6/11, -2/11

Question 4: Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

ANSWER : - The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).

It is known that the direction ratios of line joining the points, (x1y1z1) and (x2y2z2), are given by, x2 − x1y2 − y1, and z2 − z1.

The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3.

The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6.

It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional.

Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and C are collinear.

Question 5: Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2)

ANSWER : - The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the direction cosines of AB are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the direction cosines of BC are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the direction cosines of AC are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry  

 Question 7: Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

ANSWER : - Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6).

The direction ratios, a1b1c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4.

The direction ratios, a2b2c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.

AB and CD will be perpendicular to each other, if a1a2 +  b1b2c1c2 = 0

a1a2   b1b2  c1c2 = 2 × 3 5 × 2 (− 4) × 4

= 6 + 10 – 16 = 0

Therefore, AB and CD are perpendicular to each other.

Question 8: Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).

ANSWER : - Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5).

The directions ratios, a1b1c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4.

The direction ratios, a2b2c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, AB is parallel to CD.

Question 9: Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry .

ANSWER : - It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the line which passes through point A and parallel to  is given by  is a constant

Question 10: Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i - j +4k  and is in the direction i + 2j - k  .

ANSWER : - It is given that the line passes through the point with position vector

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that, a line through a point with position vector NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is given by the  equations NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Eliminating λ, we obtain the Cartesian form equation as

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 11: Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Answer: It is given that the line passes through the point (−2, 4, −5) and is parallel  to  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The direction ratios of the line NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry are 3, 5, and 6. 

The required line is parallel to  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry 

Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0 

It is known that the equation of the line through the point (x1y1z1) and with direction ratios, abc, is given by   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore the equation of the required line is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 12: The Cartesian equation of a line is   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry Write its vector form.

ANSWER : - The Cartesian equation of the line is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ...(1)

The given line passes through the point (5, −4, 6). The position vector of this point is  

Also, the direction ratios of the given line are 3, 7, and 2.

This means that the line is in the direction of vector,  

It is known that the line through position vector  and in the direction of the vector  is given by the equation  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the required equation of the given line in vector form

Question 13: Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).

ANSWER : - 

The required line passes through the origin. Therefore, its position vector is given by, a = 0 ..(1)

The direction ratios of the line through origin and (5, −2, 3) are

(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3

The line is parallel to the vector given by the equation  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The equation of the line in vector form through a point with position vector a and parallel to b is,  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The equation of the line through the point (x1y1z1) and direction ratios abc is given by,  

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the equation of the required line in the Cartesian form is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 14: Find the vector and the Cartesian equations of the line that passes through the points  (3, −2, −5), (3, −2, 6).

ANSWER : - Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its position vector is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The direction ratios of PQ are given by,

(3 − 3) = 0, (−2 2) = 0, (6 5) = 11

The equation of the vector in the direction of PQ is

The equation of PQ in Cartesian form is 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 15: Find the angle between the following pairs of lines:

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional GeometryNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - (i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(ii) The given lines are parallel to the vectors  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ,respectively.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 16: Find the angle between the following pairs of lines:

1. NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

2. NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - Let  and   be the vectors parallel to the pair of lines NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometryrespectively.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The angle, Q, between the given pair of lines is given by the relation

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(ii) Let   be the vectors parallel to the given pair of lines,  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometryrespectively.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 17: Find the values of p so the line   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry are at right angles

Answer: The given equations can be written in the standard form as

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The direction ratios of the lines are −3,2p/7,2,-3p/7,1,-5,  respectively.

Two lines with direction ratios, a1b1c1 and a2b2c2, are perpendicular to each other, if a1a2b1 b2c1c2 = 0

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

 Question 18: Show that the lines   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry are perpendicular to each other.

ANSWER : - The equations of the given lines are NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry 

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.

Two lines with direction ratios, a1b1c1 and a2b2c2, are perpendicular to each other

if a1a2b1 b2c1c2 = 0

∴ 7 × 1 + (−5) × 2 + 1 × 3

= 7 − 10 + 3

= 0

Therefore, the given lines are perpendicular to each other.

 Question 19: Find the shortest distance between the lines

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The equations of the given lines are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the shortest distance between the lines

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ..(1)

Comparing the given equations, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting all the values in equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the shortest distance between the two lines is 3√2/2 units

Question 20: Find the shortest distance between the lines  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The given lines are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the shortest distance between the two

lines,  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry , is given by.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.....(1)

Comparing the given equations, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional GeometryNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

 Substituting all the values in equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Since distance is always non-negative, the distance between the given lines is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry units.

Question 21: Find the shortest distance between the lines whose vector equations are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional GeometryNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The given lines are NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the shortest distance between the lines. NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry   , is given by.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ...(i)

Comparing the given equations with  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
obtain  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting all the values in equation (1).. we obtain 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the shortest distance between the two given lines is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry units.

  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER:-The given lilies are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the shortest distance between the lines,   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is given by.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

For the given equations,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting all the values in equation (3).. we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the shortest, distance between the lines is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry units.

Question 23: In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

(a) z = 2
(b) x + y + z = I
(c) 2x+3y-z = 5
(d) 5y+ 8 = 0

ANSWER : - (a) The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1)

The direction ratios of normal are 0, 0, and 1.

This is of the form lx +  my  + nzd, where lmn are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) xyz = 1 … (1)

The direction ratios of normal are 1, 1, and 1.

  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Dividing both sides of equation (1) by √3, , we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This equation is of the form  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry are the direction cosines of normal to the plane andd is the distance of normal from the origin.

Therefore, the direction cosines of the normal are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and the distance of normal from the origin is NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(c) 2x+3y -  z = 5 ... (1)

The direction ratios of normal are 2. 3. and-1.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Dividing both sides of equation (1) byNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This equation is of the form   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and the distance of normal from the origin is NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry units.

(d) 5y + 8 = 0

⇒ 0x- 5y + 0z = 8 ... (1)
The direction ratios of normal are 0,-5, and 0.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Dividing both sides of equation (1) by 5, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This equation is of the form lx + my + nz = d. where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are 0, -1, and 0 and the distance of normal

 Question 24: Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The normal vector is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the equation of the plane with position vector  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the vector equation of the required plane.

Question 25: Find the Cartesian equation of the following planes:

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER:-(a) It is given that equation of the plane is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

For any arbitrary point P (x.. y. z) on the plane, position vector  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
Substituting the value of  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry in equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the Cartesian equation of the plane.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

For any arbitrary point P (x, y. z) on the plane, position vector  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting the value of  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the Cartesian equation of the plane.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

For any arbitrary point P (x,y, z) on the plane, position vector  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting the value of  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the Cartesian equation of the given plane.

Question 26: In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

(a) 2x + 3y + 4z -12 = 0
(b) 3y + 4z - 6 = 0
(c) x + y+z = 1   
(d) 5 v + 8 = 0

ANSWER: - (a) Let the coordinates of the foot of perpendicular P from the origin to the plane be

(x1,y1,z1)
2x + 3 y + 4 z - 12=0
⇒ 2x + 3y + 4z = 12 ... (1)
The direction ratios of normal are 2, 3, and 4.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Dividing both sides of equation (1) by NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This equation is of the form lxmynzd, where lmn are the direction cosines of normal to the plane and d is the distance of normal from the origin

The coordinates of the foot of the perpendicular are given by (Id. md. nd).
Therefore, the coordinates of the foot of the perpendicular are NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1 z1)

3y + 4z - 6 = 0

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The direction ratios of the normal are 0; 3. and 4.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Dividing both sides of equation (1) by 5, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This equation is of the form lxmynzd, where lmn are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ldmdnd).

Therefore, the coordinates of the foot of the perpendicular are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,y1,z1)

x+y + z = 1    ... (1)
The direction ratios of the normal are 1, 1,  and 1.

  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Dividing both sides of equation (1) by √3, , we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This equation is of the form lx + my + nz = d. where l, m. n are the direction cosines of normal to the plane andd is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (Id. md. nd).
Therefore, the coordinates of the foot of the nemendicular are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,y1,z1).

5y + 8 = 0
⇒ 0x- 5y + 0z = 8 ... (1)

The direction ratios of the normal are 0, −5, and 0.

Dividing both sides of equation (1) by 5, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This equation is of the form lxmy + nzd, where lmn are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ldmdnd).

Therefore, the coordinates of the foot of the perpendicular are (0,-1,0) and (0,-8/5, 0)

Question 27: Find the vector and Cartesian equation of the planes

(a) that passes through the point (1, 0, −2) and the normal to the plane NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

(b) that passes through the point (1, 4, 6) and the normal vector to the plane

ANSWER : - (a) The position vector of point (1, 0,-2) is NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The normal vector  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry perpendicular to the plane is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The vector equation of the plane is given by,  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is the position vector of any point P (x, y. z) in the plane.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, equation (1) becomes

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the Cartesian equation of the required plane.

(b) The position vector of the point (1. 4, 6) is NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The normal vector NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry perpendicular to the plane is NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The vector equation of the plane is given by.  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is the position vector of any point P (x,y\ z) in the plane

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, equation (1) becomes

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the Cartesian equation of the required plane.

Question 28: Find the equations of the planes that passes through three points.

(a) (1, 1,-1), (6,4, -5), (-4, -2, 3)    
(b) (1,1,0), (1,2, 1), (-2,2,-1)

ANSWER: - (a) The given points are A (1,1, - 1),B (6. 4, -5), and C (-4. - 2, 3).

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

= 2+2-4
= 0

Since A,B,C are collinear points.there will be infinite number of planes passing through the given points.

(b) The given points are A (1,1,0),B (1,2, 1), andC (-2,2,-1).

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the equation of the plane through the points.  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the Cartesian equation of the required plane.

Question 29: Find the intercepts cut off by the plane 2x + y -z = 5

ANSWER:- 2x + y - z = 5    ..... (1)

Dividing both sides of equation (1) by 5, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the equation of a plane in intercept form is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry, where a, b, c are the  intercepts cut off by the plane at x,y, and z axes respectively.

Therefore, for the given equation. 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the intercepts cut off by the plane are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 30: Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

ANSWER : - The equation of the plane ZOX is

y = 0

Any plane parallel to it is of the form, ya

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

 Question 31: Find the equation of the plane through the intersection of the planes  point (2, 2, 1)

ANSWER : - The equation of any plane through the intersection of the planes,

3x − y + 2z ­− 4 = 0 and xyz − 2 = 0, is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry    .......(1)

The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation (1).

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ill equation (1). we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the required equation of the plane.

Question 32: Find the vector equation of the plane passing through the intersection of the planes NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and through the point (2.1, 3)

ANSWER : - The equations of the planes are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by;

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Hie plane passes through the point ('2,1,3). Therefore. its position vector is given by,  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting ill equation (3). we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry in equation (3), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the vector equation of the required plane.

Question 33: Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y+ 4z = 5 which is perpendicular to the plane x - y + z = 1

ANSWER : - The equation of the plane through the intersection of the planes, x + y + z = 1  and 2x + 3y + 4z = 5, is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The direction ratios, a1,b1,c1,  of this plane are (2λ + 1). (3λ+ 1), and (4λ+ 1).
The plane in equation (1) is perpendicular to x- y + z = 0
Its direction ratios, a2,b2,c2, are 1,-1, and  1.
Since the planes are perpendicular,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry in equation (1), we obtain 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the required equation of the plane. 

Question 34: Find the angle between the planes whose vector equations are NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The equations of the given planes are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that if  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry are normal to the planes. NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry , then the angle between them. Q. is given by.  

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting the value of  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ill equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 35: In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the given planes are not parallel.

The angle between them is given by,

ANSWER : - The direction ratios of normal to the plane.  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometryand  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The angle between L1 and L2 is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(a) The equations of the planes are 7x - 5y - 6z- 30=0 and

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the given planes are not perpendicular.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the given planes are not parallel.

The angle between them is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(b) The equations of the planes are NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the given planes are not perpendicular to each other.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the given planes are parallel to each other. 

(d) The equations of the planes are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the given lines are parallel to each other

(e) The equations of the given planes are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the given lines are not perpendicular to each other.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 36: In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane
(a) (0, 0, 0) 3x - 4y +12 = 3    
(b) (3, -2,1) 2x-y + 2z + 3 = 0
(c)(2,3,-5) x+2y-2z = 9    
(d)(-6,0,0) 2x-3y + bz-2 = 0

ANSWER : - It is known that the distance between a point, p(x1y1z1), and a plane, Ax+By+CzD, is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(a) The given point is (0: 0. 0) and the plane is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(b) The given point is (3, - 2,1) and the plane is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(c) The given point is (2, 3,-5) and the plane is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(d) The given point is (—6,0. 0) and the plane is NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 37: Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1), (4, 3, −1).

ANSWER : - Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1).

Also, let BC be the line joining the points, B (3, 5, −1) and C (4, 3, −1).

The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and (−1 1) = 0

OA is perpendicular to BC, if a1a2 + b1b2 + c1c2 = 0

∴ a1a2  + b1b2  + c1c2 = 2 × 1 +1 (−2) + 1 ×0 = 2 − 2 = 0

Thus, OA is perpendicular to BC.

Question 38:If l1m1n1 and l2m2n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1n2 − m2n1,n1l2 − n2l1l1m2 ­− l2m1.

ANSWER : - It is given that l1m1n1 and l2m2n2 are the direction cosines of two mutually perpendicular lines. Therefore,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction cosinesNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
l, m, n are the direction cosines of the line.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

From (1), (2).and(3), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting the values from equations (5) and (6) in equation (4), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the direction cosines of the required line are  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 39: Find the angle between the lines whose direction ratios are a. b. c and b - c. c - a. a - b.

ANSWER:-The angle O between the lines with direction cosines, a. b. c and b - c, c - a,

a - b. is given by.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the angle between the lines is 90°.

Question 40: Find the equation of a line parallel to x-axis and passing through the origin.

ANSWER : - The line parallel to x-­­axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by (a, 0, 0), where a ∈ R.

Direction ratios of OA are (a − 0) = a, 0, 0

The equation of OA is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the equation of line parallel to x-axis and pas sing through origin is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 41: If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (­−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

ANSWER : - The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (­−4, 3, −6), and

(2, 9, 2) respectively.

The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4

The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8

It can be seen that,   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, AB is parallel to CD.

Thus, the angle between AB and CD is either 0° or 180°.

Question 42: If the lilies  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry are petpendicular.find the value of k.

ANSWER : - The direction of ratios of the lines.  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry , are -3, 2k, 2  and 3k,1, - 5 respectively.


It is known that two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular,

if  a1a2 + b1b2 + c1c = 0

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, for  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry . the given lines are perpendicular to each other.

Question 43: Find the vector equation of the plane passing through (1,2, 3) and perpendicular to the plane   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The position vector of the point (1, 2, 3) is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The direction ratios of the normal to the plane.  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry .are 1.2. and -5 and the normal vector is NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The equation of a line passing through a point and perpendicular to the given plane is given by, NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 44: Find the equation of the plane passing through (a, b. c) and parallel to the plane  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - Any plane parallel to the plane.   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry , is of the form

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The plane passes through the point (a. b. c). Therefore, the position vector  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry of this point is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, equation (1) becomes

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry in equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry in equation (2), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 45: Find the shortest distance between lines NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER: - The given lines are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the shortest distance betw een two lines.  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is given by

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Comparing  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry to equations (1) and (2), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting all the values in equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the shortest distance between the two given lines is 9 units.

Question 46: Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane

ANSWER: It is known that the equation of the line passing through the points,  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The line pas sing through the points. (5. 1. 6) and (3. 4. 1), is given by.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Any point, on the line is of the form (5 - 2k. 3k + 1, 6 - 5k).

The equation of YZ-plane is x = 0

Since the line passes through YZ-plane.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the required point is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 47: Find the coordinates of the point where the line through (5,1, 6) and (3,4,1) crosses the ZX - plane.

ANSWER: - It is known that the equation of the line passing through the points,  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The line passing through the points, (5,1,6) and (3, 4,1), is given by, 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Any point, on the line is of the form  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Since the line passes through ZX-plane.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the required point is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry  

Question 48: Find the coordinates of the point where the line through (3, ­−4, −5) and (2, − 3, 1) crosses the plane 2x +  y +  z = 7).

ANSWER : - It is known that the equation of the line through the points, (x1y1z1) and (x2y2z2), is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Since the line passes through the points.  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry its equation is given by.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, any point on the line is of the form  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This point lies on the plane, 2x + y +  z = 7

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 × 2 − 5) i.e., (1, −2, 7).

Question 49: Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x +  2y +  3z = 5 and 3x +  3y +  z = 0.

ANSWER : - The equation of the plane passing through the point (−1, 3, 2) is

a (x + 1)  b (y − 3)  c (z − 2) = 0 … (1)

where, abc are the direction ratios of normal to the plane.

It is known that two planes NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry are perpendicular,  

if NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Plane (1) is perpendicular to the plane, x + 2y + 3z = 5

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Also, plane (1) is perpendicular to the plane, 3x + 3yz = 0

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

From equations (2) and (3), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting the values of ab, and c in equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the required equation of the plane.

Question 51: Find the equation of the plane passing through the line of intersection of the planes    NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and parallel to x-axis.

ANSWER : - The given planes are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

Question 52: If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.

ANSWER : - The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x1y1 z1) is

 where, a, b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 53: Find the equation of the plane which contains the line of intersection of the planes  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry & which is perpendicular to the plane  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The equations of the given planes are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ..(3)

The plane in equation (3) is perpendicular to the plane NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting l = 7/19in equation (3), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry in equation (3). 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

 Question 54: Find the distance of the point (−1, −5, −­10) from the point of intersection of the line NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry and the plane  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The equation of the given line is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ..(1)

The equation of the given plane is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ..(2)

Substituting the value of  r_ from equation (1) in equation (2), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting this value in equation (1), we obtain the equation of the line as  This means that the position vector of the point of intersection of the line & the plane is 

This shows that the point of intersection of the given line and plane is given by the coordinates,  (2, −1, 2). The point is (−1, −5, −10).

The distance d between the points, (2, −1, 2) and (−1, −5, −10), is

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry 

Question 55: Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - Let the required line be parallel to vector  b given by  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The position vector of the point (1, 2, 3) is   NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The equation of line passing through (1, 2, 3) and parallel to  b is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The equations of the given planes are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

 

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

From equations (4) and (5), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the direction ratios of NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting the value of  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ill equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the equation of the required line.

Question 56: Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines: 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Answer: - Let the required line be parallel to the vector b given by, NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The position vector of the point (1, 2, − 4) is  

The equation of the line passing through (1, 2, −4) and parallel to vector b  is 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The equations of the lines are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Line (1) and line (2) are perpendicular to each other.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Also, line (1) and line (3 ) are perpendicular to each othei

 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

From equations (4) and (5); we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

∴ Direction ratios of  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Substituting  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry ill equation (1), we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

This is the equation of the required line.

Question 57: Prove that if a plane has the intercepts abc and is at a distance of P units from the origin, then NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

ANSWER : - The equation of a plane having intercepts abc with xy, and z axes respectively is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

The distance (p) of the plane from the origin is given by,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 58:Distance between the two planes:   2x +3y + 4z = 4 and  4x+6y+8z =12  is 

(A)2 units                  

(B)4 units                  

(C)8 units

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Answer: 

The equations of the planes are    

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It can be seen that the given planes are parallel

It is known that the distance between two parallel planes, axbyczd1 and axbyczd2, is given by, 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the distance between the lines is  NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry units

Hence, the correct answer is D.

Question 59: The planes: 2x − y +  4z = 5 and 5x − 2.5y +  10z = 6 are

(A) Perpendicular               

(B) Parallel               

(C) intersect y-axis     

(C) passes through (0,0,5/4)

ANSWER : - The equations of the planes are

2x - y - 4z = 5      ... (1) 
5x - 2.5y+ 10z = 6     ... (2)

It can be seen that.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the given planes are parallel.

Hence, the correct answer is B.

Question 60: Name the octants in which the following points lie:

(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (–4, 2, 5), (–3, –1, 6), (2, –4, –7)

Answer 

The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.

The x-coordinate, y-coordinate, and z-coordinate of point (4, –2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.

The x-coordinate, y-coordinate, and z-coordinate of point (4, –2, –5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, –5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.

The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, –5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.

The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.

The x-coordinate, y-coordinate, and z-coordinate of point (–3, –1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.

The x-coordinate, y-coordinate, and z-coordinate of point (2, –4, –7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

Question 61: Find the distance between the following pairs of points:

(i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1)

(iii) (–1, 3, –4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3)

Answer 

The distance between points P(x1y1z1) and P(x2y2z2) is given by NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(i) Distance between points (2, 3, 5) and (4, 3, 1)

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

(ii) Distance between points (–3, 7, 2) and (2, 4, –1)

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 62: Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Answer 

Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively.

Points P, Q, and R are collinear if they lie on a line.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Here, PQ + QR NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry= PR

Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.

Question 63: Verify the following:

(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle.

(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right angled triangle.

(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Answer 

(i) Let points (0, 7, –10), (1, 6, –6), and (4, 9, –6) be denoted by A, B, and C respectively.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Here, AB = BC ≠ CA

Thus, the given points are the vertices of an isosceles triangle.

(ii) Let (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) be denoted by A, B, and C respectively.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, by Pythagoras theorem, ABC is a right triangle.

Hence, the given points are the vertices of a right-angled triangle.

(iii) Let (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) be denoted by A, B, C, and D respectively.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Here, AB = CD = 6, BC = AD =NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.

Therefore, ABCD is a parallelogram.

Hence, the given points are the vertices of a parallelogram.

Question 64: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer 

Let P (xy, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1).

Accordingly, PA = PB

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

⇒ x2 – 2x + 1 + y2 – 4+ 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1

⇒ –2x4y – 6z + 14 = –6x – 4y + 2+ 14

⇒ – 2x – 6z + 6x – 2z = 0

⇒ 4x – 8z = 0

⇒ x – 2z = 0

Thus, the required equation is x – 2z = 0.

Question 65: Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (–4, 0, 0) is equal to 10.

Answer 

Let the coordinates of P be (xyz).

The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively.

It is given that PA + PB = 10.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

On squaring both sides, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

On squaring both sides again, we obtain

25 (x2 + 8+ 16 + y2z2) = 625 + 16x2 + 200x

⇒ 25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 16x2 + 200x

⇒ 9x2 + 25y2 + 25z2 – 225 = 0

Thus, the required equation is 9x2 + 25y2 + 25z2 – 225 = 0.

Question 66:

Find the coordinates of the point which divides the line segment joining the points (–2, 3, 5) and (1, –4, 6) in the ratio (i) 2:3 internally, (ii) 2:3 externally.

Answer 

(i) The coordinates of point R that divides the line segment joining points P (x1y1z1) and Q (x2y2z2) internally in the ratio mare

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

Let R (x, yz) be the point that divides the line segment joining points(–2, 3, 5) and (1, –4, 6) internally in the ratio 2:3

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the coordinates of the required point areNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

(ii) The coordinates of point R that divides the line segment joining points P (x1y1z1) and Q (x2y2z2) externally in the ratio mare

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

Let R (x, yz) be the point that divides the line segment joining points(–2, 3, 5) and (1, –4, 6) externally in the ratio 2:3

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the coordinates of the required point are (–8, 17, 3).

Question 67:

Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Answer 

Let point Q (5, 4, –6) divide the line segment joining points P (3, 2, –4) and R (9, 8, –10) in the ratio k:1.

Therefore, by section formula,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, point Q divides PR in the ratio 1:2.

Question 68:

Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).

Answer 

Let the YZ planedivide the line segment joining points (–2, 4, 7) and (3, –5, 8) in the ratio k:1.

Hence, by section formula, the coordinates of point of intersection are given byNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

On the YZ plane, the x-coordinate of any point is zero.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the YZ plane divides the line segment formed by joining the given points in the ratio 2:3.

Question 67:

Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometryare collinear.

Answer 

The given points are A (2, –3, 4), B (–1, 2, 1), andNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

Let P be a point that divides AB in the ratio k:1.

Hence, by section formula, the coordinates of P are given by

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Now, we find the value of k at which point P coincides with point C.

By takingNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry, we obtain k = 2.

For k = 2, the coordinates of point P areNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

i.e., NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is a point that divides AB externally in the ratio 2:1 and is the same as point P.

Hence, points A, B, and C are collinear.

Question 68:

Find the coordinates of the points which trisect the line segment joining the points P (4, 2, –6) and Q (10, –16, 6).

Answer 
 

Let A and B be the points that trisect the line segment joining points P (4, 2, –6) and Q (10, –16, 6)

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Point A divides PQ in the ratio 1:2. Therefore, by section formula, the coordinates of point A are given by

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Point B divides PQ in the ratio 2:1. Therefore, by section formula, the coordinates of point B are given by

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, (6, –4, –2) and (8, –10, 2) are the points that trisect the line segment joining points P (4, 2, –6) and Q (10, –16, 6).

Question 69:

Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2). Find the coordinates of the fourth vertex.

Answer 

The three vertices of a parallelogram ABCD are given as A (3, –1, 2), B (1, 2, –4), and C (–1, 1, 2). Let the coordinates of the fourth vertex be D (xyz).

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

We know that the diagonals of a parallelogram bisect each other.

Therefore, in parallelogram ABCD, AC and BD bisect each other.

∴Mid-point of AC = Mid-point of BD

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

⇒ x = 1, y = –2, and z = 8

Thus, the coordinates of the fourth vertex are (1, –2, 8).

Question 70:

Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).

Answer 

Let AD, BE, and CF be the medians of the given triangle ABC.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Since AD is the median, D is the mid-point of BC.

∴Coordinates of point D =NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry= (3, 2, 0)

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
Since BE is the median. E is the mid-point of AC.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Since CF is the median, F is the mid-point of AB,

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the lengths of the medians of ΔABC areNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

Question 71:

If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of ab and c.

Answer 
 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is known that the coordinates of the centroid of the triangle, whose vertices are (x1y1z1), (x2y2z2) and (x3y3z3), areNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

Therefore, coordinates of the centroid of ΔPQR 

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is given that origin is the centroid of ΔPQR.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the respective values of ab, and c are NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 72:

Find the coordinates of a point on y-axis which are at a distance ofNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometryfrom the point P (3, –2, 5).

Answer 

If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.

Let A (0, b, 0) be the point on the y-axis at a distance of NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometryfrom point P (3, –2, 5). Accordingly, NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the coordinates of the required points are (0, 2, 0) and (0, –6, 0).

Question 73:

A point R with x-coordinate 4 lies on the line segment joining the pointsP (2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.

[Hint suppose R divides PQ in the ratio k: 1. The coordinates of the point R are given byNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Answer 

The coordinates of points P and Q are given as P (2, –3, 4) and Q (8, 0, 10).

Let R divide line segment PQ in the ratio k:1.

Hence, by section formula, the coordinates of point R are given byNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

It is given that the x-coordinate of point R is 4.

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Therefore, the coordinates of point R are

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Question 74:

If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2k2, where k is a constant.

Answer 

The coordinates of points A and B are given as (3, 4, 5) and (–1, 3, –7) respectively.

Let the coordinates of point P be (xyz).

On using distance formula, we obtain

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Now, if PA2 + PB2k2, then

NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry
NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

Thus, the required equation isNCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry.

The document NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry is a part of the JEE Course Mathematics (Maths) Class 12.
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FAQs on NCERT Solutions Class 12 Maths Chapter 11 - Three Dimensional Geometry

1. What are the basic concepts of three-dimensional geometry?
Ans. Three-dimensional geometry deals with the study of objects in three-dimensional space. The basic concepts include points, lines, planes, and solids. Points are the basic building blocks, lines are a collection of points that extend infinitely in both directions, planes are flat surfaces that extend infinitely in all directions, and solids are three-dimensional objects with length, breadth, and height.
2. How do you calculate the distance between two points in three-dimensional space?
Ans. The distance between two points (x1, y1, z1) and (x2, y2, z2) in three-dimensional space can be calculated using the distance formula: Distance = sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2) Here, sqrt denotes the square root, and ^2 represents squaring.
3. How do you find the equation of a plane in three-dimensional geometry?
Ans. To find the equation of a plane in three-dimensional geometry, you need to know a point on the plane and its normal vector. The equation of the plane can be written as: ax + by + cz + d = 0 where (x, y, z) is any point on the plane, (a, b, c) is the direction ratios of the normal vector, and d is a constant.
4. What is the concept of cross product in three-dimensional geometry?
Ans. The cross product is an operation in three-dimensional geometry that results in a vector perpendicular to the two given vectors. The cross product of two vectors A = (a1, a2, a3) and B = (b1, b2, b3) is given by: A x B = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1) This new vector is orthogonal to both A and B, and its magnitude represents the area of the parallelogram formed by the two vectors.
5. How do you determine the angle between two lines in three-dimensional space?
Ans. To determine the angle between two lines in three-dimensional space, you can use the dot product formula. Let the direction vectors of the lines be A = (a1, a2, a3) and B = (b1, b2, b3). The angle between the lines can be calculated using the formula: cos(theta) = (A · B) / (|A| |B|) Here, (A · B) represents the dot product of A and B, |A| and |B| represent the magnitudes of A and B respectively, and theta represents the angle between the lines.
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