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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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