Question

The foot of perpendicular of the point $(2,0,5)$ on the line $\frac{x+1}{2}=\frac{y-1}{5}-\frac{z+1}{-1}$ is $(a, \beta, \gamma)$. Then. which of the following is NOT correct?

• A. $\frac{\gamma}{\alpha}=\frac{5}{8}$
• B. $\frac{\beta}{\gamma}=-5$
• C. $\frac{\alpha \beta}{\gamma}=\frac{4}{15}$
• D. $\frac{\alpha}{\beta}=-8$

Answer 1

Answer: (B)

$L : \frac{ x +1}{2}=\frac{ y -1}{5}=\frac{ z +1}{-1}=\lambda$

Let foot of perpendicular is
$P (2 \lambda-1,5 \lambda+1,-\lambda-1) $
$ \overrightarrow{ PA }=(3-2 \lambda) \hat{ i }-(5 \lambda+1) \hat{ j }+(6+\lambda) \hat{ k }$
Direction ratio of line $\Rightarrow \vec{b}=2 \hat{i}+5 \hat{j}-\hat{k}$
$ \text { Now, } \Rightarrow \overrightarrow{ PA } \cdot \overrightarrow{ b }=0 $
$ \Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0 $
$ \Rightarrow \lambda=\frac{-1}{6}$
$ P (2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv P (\alpha, \beta, \gamma)$
$ \Rightarrow \alpha=2\left(-\frac{1}{6}\right)-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3} $
$ \Rightarrow \beta=5\left(-\frac{1}{6}\right)+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6}$
$ \Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6}$
$\therefore$ Check options