← 2018 Paper 1

UPSC 2018 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

Straight lines in 3D · Analytic Geometry · asked 5× in 13 yrs · Read the full method →

Question

Find the projection of the straight line x−12=y−13=z+1−1\dfrac{x-1}{2}=\dfrac{y-1}{3}=\dfrac{z+1}{-1} on the plane x+y+2z=6x+y+2z=6.

Technique

Projection of a line on a plane = intersection of the plane with the projecting plane (contains the line, ⊥\perp to the plane); normal of projecting plane =d⃗×n⃗=\vec d\times\vec n.

Solution

Step 1 — Strategy

The orthogonal projection of a line LL on a plane π\pi is the intersection of π\pi with the projecting plane π′\pi' — the plane that contains LL and is perpendicular to π\pi. So I find π′\pi', then present the projection as π∩π′\pi\cap\pi'.

Step 2 — The projecting plane π′\pi'

LL passes through P=(1,1,−1)P=(1,1,-1) with direction d⃗=(2,3,−1)\vec d=(2,3,-1). The given plane has normal n⃗=(1,1,2)\vec n=(1,1,2). The projecting plane contains both d⃗\vec d and n⃗\vec n, so its normal is

d⃗×n⃗=∣i^j^k^23−1112∣=(3⋅2−(−1)⋅1, (−1)⋅1−2⋅2, 2⋅1−3⋅1)=(7,−5,−1).\vec d\times\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&-1\\1&1&2\end{vmatrix}=(3\cdot2-(-1)\cdot1,\ (-1)\cdot1-2\cdot2,\ 2\cdot1-3\cdot1)=(7,-5,-1).

Through P=(1,1,−1)P=(1,1,-1):

7(x−1)−5(y−1)−1(z+1)=0 ⟹ 7x−5y−z−3=0.7(x-1)-5(y-1)-1(z+1)=0\ \Longrightarrow\ 7x-5y-z-3=0.

Step 3 — The projection line =π∩π′=\pi\cap\pi'

Answer

  7x−5y−z=3,x+y+2z=6.  \boxed{\;7x-5y-z=3,\qquad x+y+2z=6.\;}
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