The foot of perpendicular from $(5,7,3)$ to the join of $(9,13,15)$ and $(12,21,10)$ is
- $(-2,-19,7)$
- $(2,19,7)$
- $(2,2,3)$
- $(9,13,15)$
Solution
$ \begin{aligned} & B=(12,21,10) \\ & P=(5,7,3) \text { and } Q=(x, y, z) \end{aligned} $ Dr's of $\overleftrightarrow{A B}=(9-12,13-21,15-10)$ $ =(-3,-8,5)=(3,8,-5) $ Equation of line $\overleftrightarrow{A B}$ is $ \begin{aligned} \frac{x-9}{3} & =\frac{y-13}{8}=\frac{z-15}{-5}=\lambda \\ \frac{x-9}{3} & =\lambda, \frac{y-13}{8}=\lambda, \frac{z-15}{-5}=\lambda \end{aligned} $ $ \begin{aligned} & x=3 \lambda+9, y=8 \lambda+13, z=-5 \lambda+15 \\ & \therefore Q=(3 \lambda+9,8 \lambda+13,-5 \lambda+15) \end{aligned} $ Dr's of $\mathbf{P Q}=(3 \lambda+9-5,8 \lambda+13-7,-5 \lambda+15-3)$ Dr's $^{\prime}$ of $\mathbf{P Q}=(3 \lambda+4,8 \lambda+6,-5 \lambda+12)$ Since, $\overleftrightarrow{A B} \perp \mathbf{P Q}$ $ \begin{aligned} a_1 a_2+b_1 b_2+c_1 c_2 & =0 \\ 3(3 \lambda+4)+8(8 \lambda+6)-5(-5 \lambda+12) & =0 \\ 9 \lambda+12+64 \lambda+48+25 \lambda-60 & =0 \quad \Rightarrow \lambda=0 \end{aligned} $ Put, $\lambda=0$ in $Q$ $ \therefore \quad Q=(9,13,15) $ $\therefore$ Foot of Perpendicular $Q=(9,13,15)$ Hence, option (4) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)