NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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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 - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the direction Cosines of the line are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Thus, direction cosines of the line make an angle α with each of the coordinate axesNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the direction cosines of AB are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the direction cosines of BC are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the direction cosines of AC are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev  

 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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev .

ANSWER : - It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

It is known that, a line through a point with position vector NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev is given by the  equations NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Eliminating λ, we obtain the Cartesian form equation as

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Answer: It is given that the line passes through the point (−2, 4, −5) and is parallel  to  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The direction ratios of the line NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev are 3, 5, and 6. 

The required line is parallel to  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev 

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 - Three Dimensional Geometry JEE Notes | EduRev

Therefore the equation of the required line is

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Question 12: The Cartesian equation of a line is   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev Write its vector form.

ANSWER : - The Cartesian equation of the line is

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ...(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 - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

The equation of the line in vector form through a point with position vector a and parallel to b is,  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRevNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev and  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

The angle between the given pairs of lines is given by,   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

(ii) The given lines are parallel to the vectors  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ,respectively.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

1. NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

2. NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - Let  and   be the vectors parallel to the pair of lines NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRevrespectively.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

(ii) Let   be the vectors parallel to the given pair of lines,  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRevrespectively.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Question 17: Find the values of p so the line   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev are at right angles

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

 Question 18: Show that the lines   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev are perpendicular to each other.

ANSWER : - The equations of the given lines are NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev 

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 - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - The equations of the given lines are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is known that the shortest distance between the lines

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ..(1)

Comparing the given equations, we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

Question 20: Find the shortest distance between the lines  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - The given lines are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is known that the shortest distance between the two

lines,  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev , is given by.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev.....(1)

Comparing the given equations, we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRevNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Since distance is always non-negative, the distance between the given lines is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev units.

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRevNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - The given lines are NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev and  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is known that the shortest distance between the lines. NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev   , is given by.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ...(i)

Comparing the given equations with  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
obtain  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the shortest distance between the two given lines is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev units.

  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER:-The given lilies are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is known that the shortest distance between the lines,   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev is given by.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

For the given equations,

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the shortest, distance between the lines is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev 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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

This equation is of the form  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev 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 - Three Dimensional Geometry JEE Notes | EduRev and the distance of normal from the origin is NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

(c) 2x+3y -  z = 5 ... (1)

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Dividing both sides of equation (1) byNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev, we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

This equation is of the form   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev 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 - Three Dimensional Geometry JEE Notes | EduRev and the distance of normal from the origin is NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev units.

(d) 5y + 8 = 0

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - The normal vector is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is known that the equation of the plane with position vector  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

This is the vector equation of the required plane.

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

For any arbitrary point P (x.. y. z) on the plane, position vector  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
Substituting the value of  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev in equation (1), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

This is the Cartesian equation of the plane.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

For any arbitrary point P (x, y. z) on the plane, position vector  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting the value of  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev equation (1), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

This is the Cartesian equation of the plane.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

For any arbitrary point P (x,y, z) on the plane, position vector  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting the value of  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev equation (1), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Dividing both sides of equation (1) by NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev, we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

(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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

(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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

(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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev.

(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 - Three Dimensional Geometry JEE Notes | EduRev

The normal vector  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev perpendicular to the plane is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The vector equation of the plane is given by,  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev is the position vector of any point P (x, y. z) in the plane.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, equation (1) becomes

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

This is the Cartesian equation of the required plane.

(b) The position vector of the point (1. 4, 6) is NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The normal vector NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev perpendicular to the plane is NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The vector equation of the plane is given by.  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev is the position vector of any point P (x,y\ z) in the plane

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, equation (1) becomes

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

= 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 - Three Dimensional Geometry JEE Notes | EduRev

It is known that the equation of the plane through the points.  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

It is known that the equation of a plane in intercept form is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev, 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 - Three Dimensional Geometry JEE Notes | EduRev

Thus, the intercepts cut off by the plane are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev    .......(1)

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ill equation (1). we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev and through the point (2.1, 3)

ANSWER : - The equations of the planes are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Hie plane passes through the point ('2,1,3). Therefore. its position vector is given by,  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting ill equation (3). we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev in equation (3), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Substituting  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev in equation (1), we obtain 

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

This is the required equation of the plane. 

Question 34: Find the angle between the planes whose vector equations are NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev and  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - The equations of the given planes are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is known that if  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev are normal to the planes. NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev , then the angle between them. Q. is given by.  

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting the value of  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ill equation (1), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRevand  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The angle between L1 and L2 is given by,

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the given planes are not perpendicular.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the given planes are not parallel.

The angle between them is given by,

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

(b) The equations of the planes are NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Thus, the given planes are parallel to each other. 

(d) The equations of the planes are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Thus, the given lines are parallel to each other

(e) The equations of the given planes are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

The angle between the planes is given by,

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

(a) The given point is (0: 0. 0) and the plane is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

(b) The given point is (3, - 2,1) and the plane is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

(c) The given point is (2, 3,-5) and the plane is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

(d) The given point is (—6,0. 0) and the plane is NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction cosinesNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
l, m, n are the direction cosines of the line.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is known that

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Thus, the direction cosines of the required line are  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev are petpendicular.find the value of k.

ANSWER : - The direction of ratios of the lines.  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev , 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 - Three Dimensional Geometry JEE Notes | EduRev

Therefore, for  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev . 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 - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - The position vector of the point (1, 2, 3) is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The direction ratios of the normal to the plane.  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev .are 1.2. and -5 and the normal vector is NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The equation of a line passing through a point and perpendicular to the given plane is given by, NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Question 44: Find the equation of the plane passing through (a, b. c) and parallel to the plane  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - Any plane parallel to the plane.   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev , is of the form

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The plane passes through the point (a. b. c). Therefore, the position vector  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev of this point is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, equation (1) becomes

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev in equation (1), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev in equation (2), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Question 45: Find the shortest distance between lines NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev and  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER: - The given lines are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is known that the shortest distance betw een two lines.  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev is given by

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Comparing  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev to equations (1) and (2), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the required point is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Any point, on the line is of the form  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Since the line passes through ZX-plane.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the required point is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev  

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 - Three Dimensional Geometry JEE Notes | EduRev

Since the line passes through the points.  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev its equation is given by.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, any point on the line is of the form  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev and NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev are perpendicular,  

if NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev and  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev and parallel to x-axis.

ANSWER : - The given planes are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Question 53: Find the equation of the plane which contains the line of intersection of the planes  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev & which is perpendicular to the plane  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - The equations of the given planes are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ..(3)

The plane in equation (3) is perpendicular to the plane NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev 

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

This is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev in equation (3). 

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

 Question 54: Find the distance of the point (−1, −5, −­10) from the point of intersection of the line NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev and the plane  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - The equation of the given line is

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ..(1)

The equation of the given plane is

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ..(2)

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev 

Question 55: Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

ANSWER : - Let the required line be parallel to vector  b given by  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The position vector of the point (1, 2, 3) is   NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

The equations of the given planes are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

 

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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the direction ratios of NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting the value of  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ill equation (1), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Answer: - Let the required line be parallel to the vector b given by, NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

The equations of the lines are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

 

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

∴ Direction ratios of  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Substituting  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev ill equation (1), we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Answer: 

The equations of the planes are    

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Thus, the distance between the lines is  NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev 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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Here, PQ + QR NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev= 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 - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Here, AB = CD = 6, BC = AD =NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

⇒ 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 - Three Dimensional Geometry JEE Notes | EduRev

On squaring both sides, we obtain

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev.

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 - Three Dimensional Geometry JEE Notes | EduRev

Thus, the coordinates of the required point areNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev.

(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 - Three Dimensional Geometry JEE Notes | EduRev.

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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRevare collinear.

Answer 

The given points are A (2, –3, 4), B (–1, 2, 1), andNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev.

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 - Three Dimensional Geometry JEE Notes | EduRev

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

By takingNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev, we obtain k = 2.

For k = 2, the coordinates of point P areNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev.

i.e., NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev 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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

⇒ 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 - Three Dimensional Geometry JEE Notes | EduRev

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

∴Coordinates of point D =NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev= (3, 2, 0)

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
Since BE is the median. E is the mid-point of AC.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Thus, the lengths of the medians of ΔABC areNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev.

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 - Three Dimensional Geometry JEE Notes | EduRev

It is known that the coordinates of the centroid of the triangle, whose vertices are (x1y1z1), (x2y2z2) and (x3y3z3), areNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev.

Therefore, coordinates of the centroid of ΔPQR 

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

It is given that origin is the centroid of ΔPQR.

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Thus, the respective values of ab, and c are NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Question 72:

Find the coordinates of a point on y-axis which are at a distance ofNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRevfrom 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 - Three Dimensional Geometry JEE Notes | EduRevfrom point P (3, –2, 5). Accordingly, NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

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

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Therefore, the coordinates of point R are

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

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 - Three Dimensional Geometry JEE Notes | EduRev

Now, if PA2 + PB2k2, then

NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev
NCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev

Thus, the required equation isNCERT Solutions - Three Dimensional Geometry JEE Notes | EduRev.

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