Question

If A, B, C are points (1, 0, 4) (0, -1, 5) and (2, -3, 1) respectively, then the coordinates of the foot of the perpendicular drawn from A to the line BC are

• A. $(\frac{1}{2},-\frac{1}{2},\frac{9}{2})$
• B. $(1,-2,3)$
• C. $(\frac{3}{2},-\frac{3}{2},\frac{5}{2})$
• D. None of these

Answer 1

The correct option is D

None of these

Let M be the foot of the perpendicular drawn from the point A to the line BC. Since the point M lies on he line BC, it must divide BC in some ratiom say $\lambda :1,(\lambda \ne -1).$ Then Coordinates fo M are given by
$(\frac{0+2\lambda }{\lambda +1},\frac{-1+\lambda .(-3)}{\lambda +1},\frac{5+\lambda .1}{\lambda +1}),i.e.(\frac{2\lambda }{\lambda +1},\frac{-3\lambda -1}{\lambda +1},\frac{\lambda +5}{\lambda +1})\left(1\right)$
Now dr's of line BC are $2-0,-3(-1),1-5,i.e.2,-2,-4i.e.1,-1,-2.$
Also dr's of perpendicular line AM are
$\frac{2\lambda }{\lambda +1}-1,\frac{-3\lambda -1}{\lambda +1}-0,\frac{\lambda +5}{\lambda +1}-4i.e.\frac{\lambda -1}{\lambda +1},\frac{-3\lambda -1}{\lambda +1},\frac{1-3\lambda }{\lambda +1}.$
Now AM is $\perp$ to BC. Using the condition of perpendicularity, we get
$1\times \left(\frac{\lambda -1}{\lambda +1}\right)+(-1)\times \left(\frac{-3\lambda -1}{\lambda +1}\right)+(-2)\times \left(\frac{1-3\lambda }{\lambda +1}\right)=0\Rightarrow (\lambda -1)+(3\lambda +1)-2(1-3\lambda )=0\Rightarrow \lambda =\frac{1}{5}.$

Putting this value of $\lambda$ in (1), we get the desired foot M of the perpendicular from A as
$(\frac{2.\frac{1}{5}}{\frac{1}{5}+1},\frac{-3.\frac{1}{5}-1}{\frac{1}{5}+1},\frac{\frac{1}{5}+5}{\frac{1}{5}+1})i.e.,(\frac{1}{3},\frac{-4}{3},\frac{13}{3}).$