Question

Consider $△ABC$ formed by points $A(1,0,1),B(0,2,3)$ and $C(-1,-2,2)$ such that $AD$ is the altitude of triangle drawn from $A$ to $BC$ then equation of plane passing through origin and perpendicular to altitude $AD$ is

• A. $3x-z=0$
• B. $x-z=0$
• C. $3y-z=0$
• D. $x+y+z=0$

Answer 1

Answer: (B)

$BC:\frac{x}{1}=\frac{y-2}{4}=\frac{z-3}{1}=\lambda$
$\Rightarrow D=(\lambda ,4\lambda +2,\lambda +3)$
$\overrightarrow{AD}=(\lambda -1)\hat{i}+(4\lambda +2)\hat{j}+(\lambda +2)\hat{k}$
$\overrightarrow{AD}\cdot (\hat{i}+4\hat{j}+\hat{k})=0$
$\Rightarrow (\lambda -1)+16\lambda +8+\lambda +2=0$
$18\lambda =-9\Rightarrow \lambda =\frac{-1}{2}$
$\Rightarrow \overline{AD}=\frac{-3\hat{i}}{2}+0\hat{j}+\frac{3\hat{k}}{2}$
$\therefore \text{ plane }:-3x+3z+\lambda =0\text{ passes through }(0,0,0)$
$\lambda =0\Rightarrow x=z$