Question

The image of the foot of the perpendicular drawn from the origin to the line joining the points $(-9,4,5)$ and $(10,0,-1)$ w.r.t. $XY-$ plane will be :

• A. $(\frac{-8}{9},\frac{2}{9},\frac{-10}{9})$
• B. $(\frac{8}{9},\frac{2}{9},\frac{10}{9})$
• C. $(\frac{58}{59},\frac{112}{59},\frac{-109}{59})$
• D. $(\frac{58}{59},\frac{112}{59},\frac{109}{59})$

Answer 1

The correct option is C $(\frac{58}{59},\frac{112}{59},\frac{-109}{59})$

Equation of line joining $(-9,4,5)$ and $(10,0,-1)$ is given by,
$\frac{x+9}{-19}=\frac{y-4}{4}=\frac{z-5}{6}=k$
So any point on this line is given by, $P(-19k-9,4k+4,6k+5)$
Let origin is $O(0,0,0)$ so direction ratios of $PO$ are $P(-19k-9,4k+4,6k+5)$
For $P$ to be foot of perpendicular from origin to this line, $-19(-19k-9)+4(4k+4)+6(6k+5)=0$
$k=\frac{-31}{59}$
we get foot of the perpendicular, i.e., $(\frac{58}{59},\frac{112}{59},\frac{109}{59}).$
Now image of the point $(a,b,c)$ w.r.t $XY-$ plane is $(a,b,-c)$
$\therefore$ Required point is $(\frac{58}{59},\frac{112}{59},-\frac{109}{59}).$