Basic Concepts of Three Dimensional Geometry

# Basic Concepts of Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Table of contents Basic Concepts of 3D Geometry Direction Cosines and Ratios of a Line PLANE STRAIGHT LINE

## Basic Concepts of 3D Geometry

### Distance Between Two Points

Let P and Q be two given points in space. Let the co-ordinates of the points P and Q be  and  with respect to a set OX, OY, OZ of rectangular axes.

The position vectors of the points P and Q are given by

Now we have

### Section Formula

(for external division take –ve sign)
To determine the co-ordinates of a point R which divides the joining of two points P and  internally in the ratio m1 : m2. Let OX, OY, OZ be a set of rectangular axes.
The position vectors of the two given points  are given by

Also if the co-ordinates of the point R are (x, y, z), then

Now the point R divides the join of P and Q in the ratio m1 : m2, so that

or    [Using (1), (2) and (3)]

Remark : The middle point of the segment PQ is obtained by putting m1 = m2. Hence the co-ordinates of the middle point of PQ are

Question for Basic Concepts of Three Dimensional Geometry
Try yourself:
In a 3D space, if point P has coordinates (2, 3, 4) and point Q has coordinates (5, 1, 7), what are the coordinates of a point R that divides the line segment PQ internally in the ratio 2:3?

### Centroid:

Centroid of a Triangle :

Let ABC be a triangle. Let the co-ordinates of the vertices A, B and C be  and  respectively. Let AD be a median of the ΔABC. Thus D is the mid point of BC.

Now if G is the centroid of ΔABC, then G divides AD in the ratio 2 : 1. Let the co-ordinates of G be (x, y, z).

Centroid Of A Tetrahedron :

Let ABCD be a tetrahedron, the co-ordinates of whose vertices are  = 1, 2, 3, 4.

Let  be the centroid of the face ABC of the tetrahedron. Then the co-ordinates of  are

The fourth vertex D of the tetrahedron does not lie in the plane of ΔABC. We know from statics that the centroid of the tetrahedron divides the line DG1 in the ratio 3 : 1. Let G be the centroid of the tetrahedron and if (x, y, z) are its co-ordinates, then

Ex.1 P is a variable point and the co-ordinates of two points A and B are (–2, 2, 3) and (13, –3, 13) respectively. Find the locus of P if 3PA = 2PB.

Sol. Let the co-ordinates of P be (x, y, z).

Now it is given that 3PA = 2PB i.e., 9PA2 = 4PB2. ....(3)

Putting the values of PA and PB from (1) and (2) in (3), we get

9{(x + 2)2 + (y – 2)2 + (z – 3)2} = 4 {(x – 13)2 + (y + 3)2 + (z – 13)2}

or 9 {x2 + y2 + z2 + 4x – 4y – 6z + 17} = 4{x2 + y2 + z2 – 26x + 6y – 26z + 347}

or 5x2 + 5y2 + 5z2 + 140x – 60 y + 50 z – 1235 = 0

or x2 + y2 + z2 + 28x – 12y + 10z – 247 = 0

This is the required locus of P.

Ex.2 Find the ratio in which the xy-plane divides the join of (–3, 4, –8) and (5, –6, 4). Also find the point of intersection of the line with the plane.

Sol. Let the xy-plane (i.e., z = 0 plane) divide the line joining the points (–3,4, –8) and (5, –6, 4) in the ratio ,  in the point R. Therefore, the co-ordinates of the point R are

....(1)

But on xy-plane, the z co-ordinate of R is zero

(4μ - 8) / (μ + 1) = 0, or μ = 2. Hence  μ : 1 = 2 : 1. Thus the required ratio is 2 : 1.
Again putting  the co-ordinates of the point R become (7/3, –8/3, 0).

Ex.3 ABCD is a square of side length ‘a’. Its side AB slides between x and y-axes in first quadrant. Find the locus of the foot of perpendicular dropped from the point E on the diagonal AC, where E is the midpoint of the side AD.

Sol. Let vertex A slides on y-axis and vertex B slides on x-axis coordinates of the point A are (0, a sin θ) and that of C are (a cos θ + a sin θ, a cos θ)

...(2)

Form (1) and (2),

Question for Basic Concepts of Three Dimensional Geometry
Try yourself:
In a triangle ABC, the coordinates of vertices are A(1, 2), B(3, 4), and C(5, 6). If G is the centroid of the triangle, what are the coordinates of G?

## Direction Cosines and Ratios of a Line

### Direction Cosines

If α, β, γ are the angles which a given directed line makes with the positive direction of the axes. of x, y and z respectively, then cos a, cos β cos g are called the direction cosines (briefly written as d.c.'s) of the line. These d.c.'s are usually denoted by ℓ, m, n.

Let AB be a given line.Draw a line OP parallel to the line AB and passing through the origin O. Measure angles a, b, g, then cos a, cos β, cos g are the d.c.'s of the line AB. It can be easily seen that l, m, n, are the direction cosines of a line if and only if  is a unit vector in the direction of that line.
Clearly, OP'(i.e. the line through O and parallel to BA) makes angle 180o - α, 180- β, 180o - γ, with OX, OY and OZ respectively.

Hence d.c.'s of the line BA are cos (180o - α), cos (180o - β), cos (180o - γ) i.e., are -cos α, -cos β , - cos γ.
If the length of a line OP through the origin O be r, then the co-ordinates of P are (ℓr, mr, nr) where ℓ, m, n are the d c.'s of OP.
If ℓ, m, n are direction cosines of any line AB, then they will satisfy ℓ 2 + m2 + n2 = 1.

### Direction Ratios :

If the direction cosines ℓ, m, n of a given line be proportional to any three numbers a, b, c respectively, then the numbers a, b, c are called direction ratios (briefly written as d.r.'s of the given line.

### Relation Between Direction Cosines And Direction Ratios :

Let a, b, c be the direction ratios of a line whose d.c.’s are ℓ, m, n. From the definition of d.r.'s. we have ℓ/a = m/b = n/c = k (say). Then ℓ = ka, m = kb, n = kc.   But ℓ2 + m2 + n2 = 1.

Taking the positive value of k, we get

Again taking the negative value of k, we get

Remark. Direction cosines of a line are unique. But the direction ratios of a line are by no means unique. If a, b, c are direction ratios of a line, then ka, kb, kc are also direction ratios of that line where k is any non-zero real number. Moreover if  a, b, c are direction ratios of a line, then a ˆi + b ˆj + c kˆ is a vector parallel to that line.

Ex.4  Find the direction cosines  + m + n of the two lines which are connected by the relation  + m + n  = 0 and mn - 2n -2m = 0.
Sol.

The given relations are ℓ+ m + n = 0 or ℓ = -m - n  ....(1)
and  mn - 2nℓ - 2ℓm = 0     ...(2)

Putting the value of ℓ from (1) in the relation (2), we get mn - 2n (-m -n) - 2(-m - n) m = 0  or  2m2 + 5mn + 2n2 = 0  or   (2m + n) (m + 2n) = 0.

From (1), we have                ...(3)

Now when       ,

(3) given

∴

∴  The d.c.’s of one line are

i.e.

∴  The d.c.’s of the one line are

Again when

i.e.

∴  The d.c.’s of the other line are

To find the projection of the line joining two points P(x1, y1, z1) and Q(x2, y2, z2) on the another line whose d.c.’s are ℓ, m, n

Let O be the origin. Then

Now the unit vector along the line whose d.c.’s are ℓ,m,n

∴  projection of PQ on the line whose d.c.’s are ℓ, m, n

The angle q between these two lines is given by

If l1, m1, n1 and l2 , m2, n2 are two sets of real numbers, then

(l12 + m12 + n12) (l2+ m22 + n22) - (l1l2 + m1m+ n1n2)2 = (m1n2 – m2n1)+ (n1l- n2l1)2 + (l1m2 - l2m1)2

Now, we have

sin2 θ = 1 - cos2 θ = 1 - (l1l2 + m1m2 + n1n2)2 = (l12 + m1+ n12) (l22 + m22 + n22) - (l1l2 + m1m2 + n1n2)2

= (m1n2 – m2n1)2 + (n1l2 - n2l1)2 + (l1m2 - l2m1) =

Condition for perpendicularity

Condition for parallelism

Ex.5 Show that the lines whose d.c.’s are given by l + m + n = 0 and 2mn + 3ln - 5lm = 0 are at right angles.

Sol. From the first relation, we have l = -m - n. ..(i)

Putting this value of l in the second relation, we have

2mn + 3 (–m –n) n – 5 (–m –n) m = 0 or 5m2 + 4mn – 3n2 = 0 or 5(m/n)2 + 4(m/n) – 3 = 0 ....(2)

Let l1, m1, n1 and l2, m2, n2 be the d,c's of the two lines. Then the roots of (2) are m1/n1 and m2/n2.

product of the roots   ...(3)

Again from (1), n = – l - m and putting this value of n in the second given relation, we have

2m (–l - m) + 3l(-l - m) - 5lm = 0

or 3(l/m)2 + 10 (l/m) + 2 = 0.

From (3) and (4) we have

l1l2 + m1m2 + n1n2 = (2 + 3 - 5) k = 0 . k = 0.  ⇒  The lines are at right angles.

Remarks :

(a) Any three numbers a, b, c proportional to the directi on cosi nes are called the direction ratios

same sign either +ve or –ve should be taken throughout.

(b) If θ is the angle between the two lines whose d.c's are l 1 , m1, n1 and l 2 , m2, n2

cos θ = l1l2 + m1m2 + n1n2

Hence if lines are perpendicular then l1l2 + m1m2 + n1n2 = 0.

if lines are parallel then

Note that if three lines are coplanar then

(c) Projection of the join of two points on a line with d.c’s l, m, n are l (x2 – x1) + m(y2 – y1) + n(z2 – z1)

(d) If l1, m1, nand l2, m2, n2 are the d.c.s of two concurrent lines, show that the d.c.s of two lines bisecting the angles between them are proportional to l1 ± l2, m1 ± m2, n± n2.

Question for Basic Concepts of Three Dimensional Geometry
Try yourself:
Which of the following conditions determine that two lines are at right angles to each other based on their direction cosines?

### Area Of A Triangle

Show that the area of a triangle whose vertices are the origin and the points A(x1, y1, z1) and B( x2, y2, z2) is

The direction ratios of OA are x1, y1, z1 and those of OB are x2, y2, z2.

Also OA =

and OB =

∴  the d.c.' s of OA are

and the d.c.’s of OB are

Hence if q is the angle between the line OA and OB, then

Hence the area of ΔOAB

Ex.6 Find the area of the triangle whose vertices are A(1, 2, 3), B(2, –1, 1)and C(1, 2, –4).
Sol.
Let Δx, Δy, Δz be the areas of the projecti ons of the area Δ of triangle ABC on the yz, zx and xy-planes respectively. We have

∴ the required area

Ex.7 A plane is passing through a point P(a, –2a, 2a), a ≠ 0, at right angle to OP, where O is the origin to meet the axes in A, B and C. Find the area of the triangle ABC.

Sol.

Equation of plane passing through P(a, –2a, 2a) is A(x – a) + B(y + 2a) + C(z – 2a) = 0.

∵ the direction cosines of the normal OP to the plane ABC are proportional to a – 0, –2a – 0, 2a – 0 i.e. a, –2a, 2a.

⇒ equation of plane ABC is a(x – a) – 2a(y + 2a) + 2a(z – 2a) = 0 or ax – 2ay + 2az = 9a2 ....(1)

Now projection of area of triangle ABC on ZX, XY and YZ

planes are the triangles AOC, AOB and BOC respectively.

∴ (Area ΔABC)2 = (Area ΔAOC)2 + (Area ΔAOB)2 + (Area ΔBOC)2

Question for Basic Concepts of Three Dimensional Geometry
Try yourself:
Find the area of the triangle with vertices A(2, 3, 4), B(1, -2, 3), and C(-3, 1, 5).

## PLANE

(i) General equation of degree one in x, y, z i.e. ax + by + cz + d = 0 represents a plane.

(ii) Equation of a plane passing through (x1, y1, z1) is a(x – x1) + b (y – y1) + c(z – z1) = 0

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

(iii) Equation of a plane if its intercepts on the co-ordinate axes are

(iv) Equation of a plane if the length of the perpendicular from the origin on the plane is ‘p’ and d.c’s of the perpendiculars as ℓ , m, n is lx + my + nz = p

(v) Parallel and perpendicular planes : Two planes a1 x + b1 y + c1z + d1 = 0 and a2 x + b2 y + c2 z + d2 = 0 are

Perpendicular if a1a2 + b1b2 + c1c2 = 0, parallel if

and  Coincident if

(vi) Angle between a plane and a line is the complement of the angle between the normal to the plane and the line. If

where θ is the angle between the line and normal to the plane.

(vii) Length of the ⊥ar from a point (x1, y1, z1) to a plane ax + by + cz + d = 0 is p =

(viii) Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is

(ix) Planes bisecting the angle between two planes a1x + b1y + c1z + d1 = 0 and a2 x + b2y + c2 z + d2 = 0 is given by

of these two bisecting planes, one bisects the acute and the other obtuse angle between the given planes.

(x) Equation of a plane through the intersection of two planes P1 and P2 is given by P1 + λP2 = 0

Ex.8 Reduce the equation of the plane x + 2y – 2z – 9 = 0 to the normal form and hence find the length of the perpendicular drawn form the origin to the given plane.

Sol. The equation of the given plane is x + 2y – 2z – 9 = 0

Bringing the constant term to the R.H.S., the equation becomes x + 2y – 2z = 9 ...(1)

[Note that in the equation (1) the constant term 9 is positive. If it were negative, we would have changed the sign throughout to make it positive.]

Now the square root of the sum of the squares of the coefficients of x, y, z in (1)

Dividing both sides of (1) by 3, we have

....(2)

The equation (2) of the plane is in the normal form ℓx + my + nz = p.

Hence the d.c.’s ℓ, m, n of the normal to the plane are 1/2,2/3,-2/3   and the length p of the perpendicular from the origin to the plane is 3.

Ex.9 Find the equation to the plane through the three points (0, –1, –1), (4, 5, 1) and (3, 9, 4).

Sol. The equation of any plane passing through the point (0, –1, –1) is given by a(x – 0) + b{y – (–1)} + c{z – (–1)} = 0  or  ax + b(y + 1) + c (z + 1) = 0 ....(1)

If the plane (1) passes through the point (4, 5, 1), we have 4a + 6b + 2c = 0 ....(2)

If the plane (1) passes through the point (3, 9, 4), we have 3a + 10b + 5c = 0....(3)

Now solving the equations (2) and (3), we have

∴  a = 10λ, b = -14λ, c = 22λ.

Putting these value of a, b, c in (1), the equation of the required plane is given by

λ[10x - 14(y + 1) + 22(z + 1)] = 0  or  10x - 14(y + 1) + 22(z + 1) = 0  or 5x - 7y + 11z + 4 = 0.

Question for Basic Concepts of Three Dimensional Geometry
Try yourself:
A plane passes through the points (1, 2, 3), (4, 5, 6), and (7, 8, 9). Find the equation of the plane.

## STRAIGHT LINE

(i) Equation of a line through A(x1, y1, z1) and having direction cosines ℓ , m , n are     and the lines through (x1, y1, z1) and (x2, y2, z2)

(ii) Intersection of two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 together represent  the un symmetrical form of the straight line.

(iii) General equation of the plane containing the line   is A(x – x1) + B(y – y1) + c(z – z1) = 0 where A ℓ + bm + cn = 0.

(iv) Line of Greatest Slope AB is the line of intersection of G-plane and H is the h orizontal plane. Line of greatest slope on a given plane, drawn through a given point on the plane, is the line through the point 'P'perpendicular to the line of intersection of the given plane with any horizontal  plane.

Ex.15 Show that the di stance of the point of inter section of the line    and the plane x – y + z = 5 from the point (–1, –5, –10) is 13.
Sol.
The equation of the given line are

= r (say)   ....(1)

The co-ordinates of any point on the line (1) are (3r + 2, 4r - 1, 12 r + 2).

If this point lies on the plane x – y + z = 5, we have 3r + 2 – (4r – 1) + 12r + 2 = 5, or 11r = 0, or r = 0.

Putting this value of r, the co-ordinates of the point of intersection of the line (1) and the given plane are (2, –1, 2).

∴ The required distance = distance between the points (2, –1, 2) and (–1, –5, –10)

=

Ex.16 Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 = 0. Find also the co-ordinates of the point on the line which is at the same distance from the foot of the perpendicular as the origin is.
Sol.
The equation of the plane is 3x + 4y – 6z + 1 = 0. ....(1)

The direction ratios of the normal to the plane (1) are 3, 4, –6.

Hence the line normal to the plane (1) has d.r.’s 3, 4, –6, so that the equations of the line through (0, 0, 0) and perpendicular to the plane (1) are x/3 = y/4 = z/–6 = r (say) ....(2)

The co-ordinates of any point P on (2) are  (3r, 4r, – 6r) ....(3) If this point lies on the plane (1), then 3(3r) + r(4r) – 6(–6r) + 1 = 0, or r = –1/61.

Putting the value of r in (3), the co-ordinates of the foot of the perpendicular P are (–3/61, –4/61, 6/61).

Now let Q be the point on the line which is at the same distance from the foot of the perpendicular as the origin. Let (x1, y1, z1) be the co-ordinates of the point Q. Clearly P is the middle point of OQ.
Hence we have

or x1 = 6/61, y1 = –8/61, z1 = 12/61.

∴ The co-ordinates of Q are (–6/61, –8/61, 12/61).

Ex.17 Find in symmetrical form the equations of the line 3x + 2y – z – 4 = 0 & 4x + y – 2z + 3 = 0 and find its direction cosines.

Sol. The equations of the given line in general form are 3x + 2y – z – 4 = 0 & 4x + y – 2z + 3 = 0  ..(1) Let l, m, n be the d.c.s of the line. Since the line is common to both the planes, it is perpendicular to the normals to both the planes. Hence we have 3l + 2m – n = 0, 4l + m – 2n = 0.

Solving these, we get

or

∴ the d.c.’s of the line are

Now to find the co-ordinates of a point on the line given by (1), let us find the point where it meets the plane z = 0.

Putting z = 0 i the equations given by (1), we have 3x + 2y – 4 = 0, 4x + y + 3 = 0.

Solving these, we get  , or x = –2, y = 5.
Therefore the equation of the given line in symmetrical form is

Ex.18 Find the equation of the plane through the line 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 and parallel to the line x = 2y = 3z.

Sol. The equation of the given line are 3x – 4y + 5z = 10, 2x + 2y – 3z = 4 ...(1)

The equation of any plane through the line (1) is (3x – 4y + 5z – 10) + l (2x + 2y - 3z - 4) = 0

or  (3 + 2λ)x + (-4 +2λ) y + (5 - 3λ) z - 10 - 4λ = 0. ...(2)

The plane (1) will be parallel to the line x = 2y = 3z

(3 + 2λ) . 6 + (-4 + 2λ). 3 + (5 - 3λ).2 = 0  or  λ(12 + 6 - 6) + 18 - 12 + 10 = 0 or λ =-4/3

Putting this value of l in (2), the required equation of the plane is given by

or x – 20y + 27z = 14.

Ex.19 Find the equation of a plane passing through the line   and making an angle of 30° with the plane x + y + z = 5.

Sol. The equation of the required plane is (x – y + 1) + λ (2y + z - 6) = 0 Þ x + (2l - 1) y + λz + 1 - 6l = 0

Since it makes an angle of 30° with x +y + z = 5

are two required planes.

Ex.20 Prove that the lines 3x + 2y + z – 5 = 0 = x + y – 2z – 3 and 2x – y – z = 0 = 7x + 10y – 8z – 15 are perpendicular.

Sol. Let l1, m1, n1 be the d.c.s of the first line. Then 3l1 + 2m1 + n1 = 0, l1 + m1 - 2n1 = 0. Solving, we get

Again let l2, m2,n2 be the d.c.s of the second line, then 2l2 - m2 - n2 = 0, 7l2 + 10m2 - 8n2 = 0.

Hence the d.c.’s of the two given lines are proportional to –5, 7, 1 and 2, 1, 3.

We have –5.2 + 7.1 + 1.3 = 0

Therefore, the given lines are perpendicular.

The document Basic Concepts of Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
All you need of JEE at this link: JEE

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

## FAQs on Basic Concepts of Three Dimensional Geometry - Mathematics (Maths) for JEE Main & Advanced

 1. What are the direction cosines of a line in three-dimensional geometry?
Ans. The direction cosines of a line in three-dimensional geometry are the cosines of the angles that the line makes with the positive x, y, and z-axes. They provide a way to describe the orientation of a line in space.
 2. How are direction ratios related to direction cosines in three-dimensional geometry?
Ans. The direction ratios of a line in three-dimensional geometry are proportional to the direction cosines. Specifically, the direction ratios are obtained by dividing the direction cosines by a common factor to simplify the representation of the line's direction.
 3. How can the equation of a plane be represented in three-dimensional geometry?
Ans. The equation of a plane in three-dimensional geometry can be represented in the form Ax + By + Cz = D, where A, B, and C are the direction cosines of the normal to the plane, and D is a constant term determined by the position of the plane in space.
 4. What is the significance of the concept of a straight line in three-dimensional geometry?
Ans. In three-dimensional geometry, a straight line serves as a fundamental geometric object that connects two points in space. Understanding the properties and equations of a straight line is essential for solving problems related to distance, intersection, and orientation in three dimensions.
 5. How do the concepts of planes and straight lines intersect in three-dimensional geometry?
Ans. The intersection of a plane and a straight line in three-dimensional geometry can result in various scenarios, such as a single point of intersection, no intersection (parallel), or an entire line of intersection. Understanding how planes and straight lines interact is crucial for solving complex geometric problems in three dimensions.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

### Up next

 Explore Courses for JEE exam
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;