Welcome to
3D Geometry
● Three coordinate axes divide
space into 8 octantants.
● Any coordinate on x-axis = (𝑥, 0,0)
● Any coordinate on y-axis = (0, 𝑦, 0)
● Any coordinate on z-axis = (0,0, 𝑧)
● Any coordinate on xy-plane = (𝑥, 𝑦, 0)
● Any coordinate on yz-plane = (0, 𝑦, 𝑧)
● Any coordinate on zx-plane = (𝑥, 0, 𝑧)
● Equation of xy-plane is 𝑧 = 0
● Equation of yz-plane is 𝑥 = 0
● Equation of zx-plane is 𝑦 = 0
Sign of coordinate points
Octants 𝐼 𝐼𝐼 𝐼𝐼𝐼 𝐼𝑉 𝑉 𝑉𝐼 𝑉𝐼𝐼 𝑉𝐼𝐼𝐼
Coordinates
𝑥 + − − + + − − +
𝑦 + + − − + + − −
𝑧 + + + + − − − −
𝑋′
𝑃(𝑥, 𝑦, 𝑧)
𝑌′ 𝑌
𝑂
𝑍′
Three-Dimensional Geometry
𝑍
● It is a geometric setting, in which three different
parameters (dimensions) 𝑥, 𝑦, 𝑧 are required to
determine position of a point.
𝑃(𝑥, 𝑦, 𝑧)
𝑌
Three-Dimensional Geometry
𝑍
● To reach at the point P(3,4,5) we need to move 3
units along 𝑥 − axis, 4 units along
y − axis parallelly and 5 units along z − axis parallelly
(0,0,5) (3,0,5)
● Projection of (3,4,5) on 𝑥 − axis is (3,0,0).
(0,4,5) 𝑃(3,4,5) ● Projection of (3,4,5) on y − axis is (0,4,0).
● Projection of (3,4,5) on z − axis is (0,0,5).
(3,0,0) 𝑋
● Projection of (3,4,5) on 𝑥𝑦 plane is (3,4,0).
(0,4,0) (3,4,0) ● Projection of (3,4,5) on 𝑦𝑧 plane is (0,4,5).
𝑌 ● Projection of (3,4,5) on 𝑥𝑧 plane is (3,0,5).
Distance formula between two points
● Distance = 𝑃𝑄 = 𝑥1 − 𝑥2 2 + 𝑦1 − 𝑦2 2 + 𝑧1 − 𝑧2 2
𝑌
𝑄 𝑥2 , 𝑦2 , 𝑧2
𝑃 𝑥1 , 𝑦1 , 𝑧1
𝑍
Distance of coordinate axis
● Distance of 𝑃 from 𝑥−axis = 𝑃𝐴
𝑌 𝑃𝐴 = 𝑦 2 + 𝑧 2
𝐵(0, 𝑦, 0) Similarly,
𝑃(𝑥, 𝑦, 𝑧) ● Distance of 𝑃 from 𝑦−axis = 𝑃𝐵 = 𝑥 2 + 𝑧 2
𝐴(𝑥, 0,0) ● Distance of 𝑃 from 𝑧−axis = 𝑃𝐶 = 𝑥 2 + 𝑦 2
𝑋
𝑂
𝐶(0,0, 𝑧)
𝑅(𝑥, 𝑦, 0)
𝑍 𝑍
Projection of the point 𝑃(−3, 4, −5) on 𝑦 − axis is :
A −3, 0, 0
B 0, 4, −5
C 0, 4, 0
D 0, 0, −5
Projection of the point 𝑃(−3, 4, −5) on 𝑦 − axis is :
Projection of a point on the on the y – axis is the y – coordinate of that
point itself.
Therefore, projection of 𝑃(−3, 4, −5) on 𝑦 − axis is (0, 4, 0).
A −3, 0, 0
B 0, 4, −5
C 0, 4, 0
D 0, 0. −5
Find the distance of 𝑎, 𝑏, 𝑐 from its projection on 𝑥-axis.
Projection of point 𝑎, 𝑏, 𝑐 on 𝑥-axis = 𝑎, 0, 0
Thus, the required distance is between the points 𝑎, 𝑏, 𝑐 and 𝑎, 0, 0
As a result, required distance d = 𝑎−𝑎 2 + 𝑏−0 2 + 𝑐−0 2 = 𝑏2 + 𝑐 2
Find the distance of point 𝑎, 𝑏, 𝑐 from its projection on 𝑦𝑧 plane
Projection of point 𝑎, 𝑏, 𝑐 on yz − plane = 0, 𝑏, 𝑐
Thus, the required distance is between the points 𝑎, 𝑏, 𝑐 and 0, 𝑏, 𝑐
As a result, required distance d = 𝑎−0 2 + 𝑏−𝑏 2 + 𝑐−𝑐 2 = 𝑎2 = |𝑎|
Planes are drawn parallel to the coordinate planes through the
points 1, 2, 3 and 2, 4, 7 . Find the length of edges of cuboid so
formed.
A 2, 3, 4
B 1, 2, 4
C 4, 3, 2
D 3, 2, 4
Planes are drawn parallel to the coordinate planes through the
points 1, 2, 3 and 2, 4, 7 . Find the length of edges of cuboid so
formed.
𝑃𝐶 = 2 − 1 = 1 𝑌
𝐸 𝐹 A 2, 3, 4
𝑃𝐸 = 4 − 2 =2
𝑃𝐵 = 7 − 3 =4 B 1, 2, 4
𝑄(2, 4, 7) 𝑂 𝑋
∴ Length of edges are 1, 2, 4. 𝐷 𝐶
𝑃(1, 2, 3)
C 4, 3, 2
𝐵 𝐴 D 3, 2, 4
𝑍
In 3-D, if 𝑥 ⋅ 𝑦 = 0, then it represents,
A. 2 ⊥ lines B. 2 ⊥ planes
C. 2 ∥ planes D. 2 intersecting planes
𝑥. 𝑦 = 0
⇒ 𝑥 = 0 ⇒ 𝑦𝑧 − plane
Or y = 0 ⇒ 𝑧𝑥 − plane
And as we know, 𝑦𝑧 − plane and 𝑧𝑥 − plane are
perpendicular to each other
Therefore, 𝑥. 𝑦 = 0 represents 2 ⊥ planes.
𝑥 2 + 𝑦 2 = 0 represents,
A. 2 ∥ planes B. 2 ⊥ planes
C. 2 ∥ lines D. Straight line
𝑥2 + 𝑦2 = 0
⇒ 𝑥 = 0 and y = 0
⇒ 𝑧 − axis
Therefore, 𝑥 2 + 𝑦 2 = 0 represents a straight line ( 𝑧 − axis )
As a result, D. is the correct option.
If the sum of the squares of the distances of a point from the
three coordinate axes be 36, then its distance from origin is :
A 6 units
B 3 2 units
C 6 2 units
D 2 3 units
If the sum of the squares of the distances of a point from the
three coordinate axes be 36, then its distance from origin is :
𝑌
Let 𝑃 ≡ (𝑥, 𝑦, 𝑧)
A 6 units
𝑃𝐴2 + 𝑃𝐵2 + 𝑃𝐶 2 = 36
𝐵(0, 𝑦, 0)
𝑃𝐴2 = 𝑦 2 + 𝑧 2 𝑃𝐵2 = 𝑥 2 + 𝑧 2 𝑃𝐶 2 = 𝑥 2 + 𝑦 2
𝑃(𝑥, 𝑦, 𝑧) B 3 2 units
2 2 2 2 2 2
𝑦 +𝑧 + 𝑥 +𝑧 + 𝑥 +𝑦 = 36
𝐴(𝑥, 0,0)
⇒ 2 𝑥 2 + 𝑦 2 + 𝑧 2 = 36 𝑂 𝑋
𝐶(0,0, 𝑧) C 6 2 units
2 2 2
⇒ 𝑥 +𝑦 +𝑧 = 18
𝑍
𝑟Ԧ position vector of 𝑃
D 2 3 units
⇒ 𝑂𝑃 = 𝑥2 + 𝑦2 + 𝑧2
⇒ 𝑂𝑃 = 3 2
The locus of a point 𝑃 which moves such that 𝑃𝐴2 − 𝑃𝐵2 = 5,
where 𝐴 and 𝐵 are 3, 4, 5 and −1, 3, −7 respectively, is :
A 8𝑥 + 2𝑦 + 24𝑧 − 9 = 0
B 8𝑥 + 2𝑦 + 24𝑧 + 14 = 0
C 8𝑥 − 2𝑦 + 24𝑧 − 5 = 0
D 8𝑥 − 2𝑦 − 24𝑧 + 13 = 0
The locus of a point 𝑃 which moves such that 𝑃𝐴2 − 𝑃𝐵2 = 5,
where 𝐴 and 𝐵 are 3, 4, 5 and −1, 3, −7 respectively, is :
Let 𝑃 ≡ (𝑥, 𝑦, 𝑧)
A 8𝑥 + 2𝑦 + 24𝑧 − 9 = 0
𝑃𝐴2 = 𝑥 − 3 2
+ 𝑦−4 2
+ 𝑧−5 2
𝑃𝐵2 = 𝑥 + 1 2
+ 𝑦−3 2
+ 𝑧+7 2 B 8𝑥 + 2𝑦 + 24𝑧 + 14 = 0
𝑃𝐴2 − 𝑃𝐵2 = 5 C 8𝑥 − 2𝑦 + 24𝑧 − 5 = 0
2 2 2
⇒ 𝑥−3 + 𝑦−4 + 𝑧−5
D 8𝑥 − 2𝑦 − 24𝑧 + 13 = 0
2 2 2
− 𝑥+1 + 𝑦−3 + 𝑧+7 =5
⇒ −8𝑥 + 8 − 2𝑦 + 7 − 24𝑧 − 24 = 5
Locus of P ∶ 8𝑥 + 2𝑦 + 24𝑧 + 14 = 0
Section formula
● Coordinate of a point 𝑀 which divides the line
segment joining points 𝑃 & 𝑄 in 𝑚 ∶ 𝑛, is :
𝑀 ≡ 𝑥, 𝑦, 𝑧 𝑌
𝑚𝑥2 + 𝑛𝑥1 𝑚𝑦2 + 𝑛𝑦1 𝑚𝑧2 + 𝑛𝑧1 𝑄 𝑥2 , 𝑦2 , 𝑧2
≡ , ,
𝑚+𝑛 𝑚+𝑛 𝑚+𝑛
𝑀 𝑥, 𝑦, 𝑧
𝑃 𝑥1 , 𝑦1 , 𝑧1
𝑋
𝑍
If a point 𝑅 4, 𝑦, 𝑧 lies on the line segment joining the points
𝑃(2, −3, 4) and 𝑄 8, 0, 10 , then the distance of 𝑅 from origin is :
A. 6 B. 53
C. 2 14 D. 2 21 JEE MAIN 2019
𝑃𝑅 𝜆
Let = (internally)
𝑅𝑄 1
𝑚𝑥2 +𝑛𝑥1 𝑚𝑦2 +𝑛𝑦1 𝑚𝑧2 +𝑛𝑧1 𝑌
𝑅 ≡ 𝑥, 𝑦, 𝑧 ≡ 𝑚+𝑛
,
𝑚+𝑛
,
𝑚+𝑛
𝑄 8, 0, 10
8𝜆+2 0+ −3 10𝜆+4
≡ , ,
𝜆+1 𝜆+1 𝜆+1
𝑅 4, 𝑦, 𝑧
8𝜆+2 1
4= ⇒𝜆=
𝜆+1 2
𝑃 2, −3, 4
−3 𝑂
𝑦= = −2 𝑋
𝜆+1
10𝜆+4
𝑧= =6
𝜆+1
⇒ 𝑅 = 4, −2,6 ⇒ 𝑂𝑅 = 2 14 𝑍
Find the coordinates of the points which trisect the line
segment joining t he points 𝑃 4,2, −6 and Q 10, −16,6 .
Let 𝐴 and 𝐵 be the points that trisect the line segment joining the
𝑃 4,2, −6 and Q 10, −16,6 .
Point 𝐴 divides 𝑃𝑄 in the ratio 1: 2
Therefore, by section formula, the coordinates of point 𝐴 are given by
1 × 10 + 2 × 4 1 × −16 + 2 × 2 1 × 6 + 2 × −6
, , = (6, −4, −2)
1+2 1+2 1+2
Point 𝐵 divides 𝑃𝑄 in the ratio 2: 1
Therefore, by section formula, the coordinates of point 𝐵 are given by
2 × 10 + 1 × 4 2 × −16 + 1 × 2 2 × 6 + 1 × −6
, , = (8, −10,2)
1+2 1+2 1+2
Thus, (6, −4, −2) and (8, −10,2) are the points that trisect the line
segment joining the points 𝑃 4,2, −6 and Q 10, −16,6 .
The centroid of a triangle ABC is at the point 1, 1, 1 . If the coordinates
of 𝐴 and 𝐵 are 3, – 5, 7 and – 1, 7, – 6 , respectively, find the coordinates
of the point 𝐶.
Let the coordinates of 𝐶 be 𝑥, 𝑦, 𝑧 and the
coordinates of the centroid 𝐺 be 1, 1, 1 . Then
𝑥+3−1
, i.e., 𝑥 = 1;
3
𝑦−5+7
, i.e., 𝑦 = 1;
3
𝑥+7−6
, i.e., 𝑧 = 2.
3
Hence, coordinates of 𝐶 are 1, 1, 2 .
Find the equation of set of points 𝑃, sum of whose distances
from 𝐴 4,0,0 and 𝐵 −4,0,0 is equal to 10.
Let the point 𝑃(𝑥, 𝑦, 𝑧)
Then, 𝑃𝐴 = 𝑥−4 2 + 𝑦 2 + 𝑧 2 and 𝑃𝐵 = 𝑥+4 2 + 𝑦2 + 𝑧2
Given 𝑃𝐴 + 𝑃𝐵 = 10 ⇒ 𝑃𝐴 = 10 − 𝑃𝐵
⇒ 𝑃𝐴2 = 10 − 𝑃𝐵 2
2
⇒ 𝑥−4 + 𝑦 2 + 𝑧 2 = 10 − 𝑃𝐵 2
⇒ 𝑥 2 + 16 − 8𝑥 + 𝑦 2 + 𝑧 2 = 100 + 𝑃𝐵2 − 20𝑃𝐵
⇒ 𝑥 2 + 16 − 8𝑥 + 𝑦 2 + 𝑧 2 = 100 + 𝑥 + 4 2
+ 𝑦 2 + 𝑧 2 − 20𝑃𝐵
⇒ 20𝑃𝐵 = 16𝑥 + 100 ⇒ 5𝑃𝐵 = 4𝑥 + 25
2 2
⇒ 5𝑃𝐵 = 4𝑥 + 25
2
⇒ 25{ 𝑥 + 4 + 𝑦 2 + 𝑧 2 } = 16𝑥 2 + 625 + 200𝑥
⇒ 25(𝑥 2 + 16 + 8𝑥 + 𝑦 2 + 𝑧 2 ) = 16𝑥 2 + 625 + 200𝑥
⇒ 9𝑥 2 + 25𝑦 2 + 25𝑧 2 − 225 = 0
Thus, the required equation is 9𝑥 2 + 25𝑦 2 + 25𝑧 2 − 225 = 0
