Question

Consider $\triangle ABC$ 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 $A D$ is

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

Answer 1

Answer: (B)

$B C: \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 }:-3 x +3 z +\lambda=0 \text { passes through }(0,0,0) $
$\lambda=0 \Rightarrow x = z $