Question

The coordinates of the foot perpendicular from the point $(1,0,0)$ to the line
$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ are

• A. $(2,-3,8)$
• B. $(1,-1,-10)$
• C. $(5,-8,-4)$
• D. $(3,-4,-2)$

Answer 1

Answer: (D)

Let the equation of AB is
$\frac{x-1}{2}=\frac{y-\left(-1\right)}{-3}=\frac{z-\left(-10\right)}{8}=k$
Let L be the foot of the perpendicular drawn form P$\left(1,0,0\right)$.
$\therefore L=\left(2k+1,-3k-1,8k-10\right)$.
Now, direction ratio of PL $=\left(2k,-3k-1,8k-10\right)$ and direction ratio of AB $=\left(2,-3,8\right)$ Since, PL is perpendicular to AB
$\therefore 2\left(2k\right)-3\left(-3k-1\right)+8\left(8k-10\right)=0$
Now, $k=\frac{2\left(1-1\right)+\left(-3\right)\left(0+1\right)+8\left(0+10\right)}{{\left(2\right)}^2+{\left(-3\right)}^2+{\left(8\right)}^2}$
$=\frac{0-3+80}{4+9+64}=\frac{77}{77}=1$
$\therefore$ Required co-ordinate
$=Z=\left(2+1,-3-1,8-10\right)=\left(3,-4,-2\right).$