Question

The foot of the perpendicular drawn from the point$(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane:

• A. $x-y-2z=1$
• B. $x-2y+z=1$
• C. $2x+y-z=1$
• D. $x+2y-z=1$

Answer 1

The correct option is C

$2x+y-z=1$

Explanation for correct option:

Calculate the foot of perpendicular

Foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane.

According to given condition,

Equation of $AB=\overrightarrow{r}=\left(\hat{i}+\hat{j}\right)+\lambda \left(3\hat{j}-3\hat{k}\right)$

Let , coordinated of $M=\left(1,\left(1+3\lambda \right),-3\lambda \right)$

$$\begin{gathered} \overrightarrow{PM}=-3\hat{i}+\left(3\lambda -1\right)\hat{j}-3\left(\lambda +1\right)\hat{k} \\ \overrightarrow{AB}=3\hat{j}-3\hat{k} \end{gathered}$$

$$\begin{gathered} \text{∵P}\overrightarrow{\mathrm{M}}\perp \overrightarrow{AB} \\ \Rightarrow \mathrm{P}\overrightarrow{\mathrm{M}}.\overrightarrow{AB}=0 \\ \Rightarrow 3(3\lambda -1)+9(\lambda +1)=0 \\ \Rightarrow 9\lambda -3+9\lambda +9=0 \\ \Rightarrow \lambda =-\frac{1}{3} \end{gathered}$$

$\Rightarrow \hspace{0.17em}M=(1,0,1)$ is on $2x+y-z=1$. $\left[\because M(1,(1+3\lambda ),-3\lambda )\right]$

Hence, the foot of the perpendicular lies on the plane $2x+y-z=1$.

Therefore, option (C) is the correct answer.