UPSC 2024 Maths Optional Paper 1 Q2c — Step-by-Step Solution

20 marks · Section A

Plane · Analytic Geometry · asked 5× in 14 yrs · Read the full method →

Question

Find the image of the line $x=3-6t,\;y=2t,\;z=3+2t$ in the plane $3x+4y-5z+26=0$ .

Technique

Find where the line meets the plane (this point is its own reflection); reflect the direction vector in the plane; write the image line through that point with the reflected direction.

Solution

$The original line \ell (red) through P_0=(3,0,3) pierces the plane \pi at P_1=(-3,2,5), which is fixed under reflection. The image line \ell' (blue) passes through the same point P_1 with the reflected direction; P_0'=(\tfrac35,-\tfrac{16}{5},7) is the mirror image of P_0, and P_1\to P_0' confirms the reflected direction (-9,13,-5).$

Step 1 — Identify line and plane.

Direction of line: $\vec d=(-6,2,2)$ (or simplified $(-3,1,1)$ ). A point on the line: $P_0=(3,0,3)$ at $t=0$ .

Plane normal: $\vec n=(3,4,-5)$ , $|\vec n|^2=50$ .

Step 2 — Intersection of the line with the plane.

Substitute the line into $3x+4y-5z+26=0$ :

3(3-6t)+4(2t)-5(3+2t)+26=9-18t+8t-15-10t+26=20-20t=0\;\Rightarrow\;t=1.

The intersection point $P_1=(-3,2,5)$ lies on the plane and is its own image.

Step 3 — Reflect the direction $\vec d$ in the plane.

\vec d'=\vec d-2\frac{\vec d\cdot\vec n}{|\vec n|^2}\vec n,\qquad \vec d\cdot\vec n=(-3)(3)+(1)(4)+(1)(-5)=-10.

\vec d'=(-3,1,1)-2\cdot\frac{-10}{50}(3,4,-5)=(-3,1,1)+\tfrac{2}{5}(3,4,-5)=\Bigl(-\tfrac{9}{5},\tfrac{13}{5},-1\Bigr).

Multiply by $5$ : $\vec d'=(-9,13,-5)$ .

Step 4 — Image line.

Through $P_1=(-3,2,5)$ with direction $(-9,13,-5)$ :

Answer

\boxed{\;\frac{x+3}{-9}=\frac{y-2}{13}=\frac{z-5}{-5}.\;}