Test: 3-D Geometry: Section Formula(12 Dec) - JEE MCQ

Centroid of a triangle with vertices (x₁,y₁);(x₂ ,y₂) and (x₃ ,y₃) is calculated by the formula (x₁+x₂+x₃)/3 , (y₁+y₂+y₃​)/3, (z₁+z₂+z₃​)/3
So, centroid =[(2+3-2)/3 , (-1+3+1)/3 , (6-2-1)/3]
= [1,1,1]

Let the YZ plane divide the line segment joining points (2,4,5) and (3,5,7) in the ratio k:1.
Hence, by section formula, the coordinates of point of intersection are given by :
[k(3)+2]/(k+1), [k(5)+4]/(k+1), [k(8)+7]/(k+1)
On the YZ plane, the x-coordinate of any point is zero.
(3k+2)/(k+1) = 0
3k + 2 = 0
k = -2/3

The coordinates of points P and Q are given as P(2,−3,5) and (7,0,-1)
Let R divide line segment PQ in the ratio k:1
Hence by section formula, the coordinates of point R are given by,
(k(7)+2/k+1,k(0)−3/k+1, k(-1)+5/k+1)
=(7k+2/k+1, −3/k+1, -1k+5/k+1)
It is given that the x-coordinate of point R is 1.
∴ 7k+2/k+1=1
⇒ 7k+2=k+1
⇒ 6k=-1
⇒ k=-1/6
Hence the coordinates of R are (1,2,3).

The coordinates of the centroid of △ABC
=[(2a−8+4)/3 , (3b+14+2)0/3 , (6−10+2c)/3]
=[(2a-4)/3 , (3b+16)/3 , (2c−4)/3]​
It is given that origin is the centroid of △ABC
∴ (0,0,0)=[(2a+4)/3 , (3b+16)/3 , (2c−4)/3]
(2a+4)/3 = 0 , (3b+16)/3 = 0and (2c−4)/3 = 0
⇒ a=−2 and c=2

The coordinates of the centroid of △PQR
=[(2a−4+8)/3 , (2+3b+14)/3 , (6−10+2c)/3] =((2a+4)/3, (3b+16)/3 ,(2c−4)/3)
It is given that origin is the centroid of △PQR
∴ (0,0,0)=((2a+4)/3, (3b+16)/3, (2c−4)/3)
⇒ (2a+4)/3 =0, (3b+16)/3 = 0 and (2c−4)/3=0
⇒ a =−2,b =−16/3 and c = 2

Let P be the point where the line joining the given two points (1,−2,3) and (4,2,−1) intersects the X−Y plane in m:n ratio. We are to find m:n.
Now the co-ordinate of the point P be [(4m+n)/m+n , (2m−2n)/m+n , (−m+3n)/m+n)].
As the point P lies on the X−Y plane, (−m+4n)/m+n = 0
or, −m+3n=0
or, m/n = 3/1
or, m:n = 3:1

Using section formula
 The coordinates of point R that divides the line segment joining points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) externally in the ratio m: n are .