Question

$\Delta ABC$ is formed by $A(1,8,4),B(0,-11,4)$ and $C(2,-3,1).$ If $D$ is the foot of the perpendicular from $A$ to $BC$. Then the coordinates of $D$ are

• A. (-4,5,2)
• B. (4,5,-2)
• C. (4,-5,2)
• D. (4,-5,-2)

Answer 1

Answer: (B)

Equation of line $BC$ is
$\frac{x}{2}=\frac{y+11}{8}=\frac{z-4}{-3}=\lambda$ (say)

Thus, coordinates of any point on the line $BC$ are
$(2\lambda ,8\lambda -11,-3\lambda +4)$
$\therefore$ Coordinates of $D$ are of form
$(2\lambda ,8\lambda -11,-3\lambda +4)$
Now, DR's of $BC$ are $2,8,-3$
and DR's of $AD$ are $2\lambda -1,8\lambda -19,-3\lambda$
$\because AD\perp BC$
$\therefore 2(2\lambda -1)+8(8\lambda -19)+(-3)(-3\lambda )=0$
$\Rightarrow 4\lambda -2+64\lambda -152+9\lambda =0$
$\Rightarrow 77\lambda =154$
$\Rightarrow \lambda =2$
Hence, the coordinates of $D$ are $(4,5,-2)$.