Aspire Faculty ID #17306 · Topic: CUET UG 2022 Applied Mathematics · Just now
CUET UG 2022 Applied Mathematics

A line

$\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-1}{-1}$

is perpendicular to a plane $(P)$ which passes through the point $(4,3,9)$.

If the mirror image of point $S$ on the line $(L)$ in the given plane $(P)$ is $(2,3,1)$, then coordinates of point $S$ is:

Solution

Direction ratios of line

$\vec n=(1,2,-1)$

Since the line is perpendicular to the plane, this vector is the **normal vector of the plane**.

Plane passing through $(4,3,9)$

Equation

$1(x-4)+2(y-3)-1(z-9)=0$

$x+2y-z-1=0$


Mirror image property:

The plane is the midpoint between $S$ and its image.

Image point

$S'=(2,3,1)$


Parametric form of line

$x=2+t$

$y=3+2t$

$z=1-t$

So any point on the line

$S=(2+t,3+2t,1-t)$


Midpoint between $S$ and $S'$

$\left(\frac{2+t+2}{2},\frac{3+2t+3}{2},\frac{1-t+1}{2}\right)$

$=\left(2+\frac{t}{2},3+t,1-\frac{t}{2}\right)$


Substitute into plane

$x+2y-z-1=0$

$\left(2+\frac{t}{2}\right)+2(3+t)-\left(1-\frac{t}{2}\right)-1=0$

$6+3t=0$

$t=-2$


Hence

$S=(2-2,3-4,1+2)$

$S=(0,-1,3)$

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