Question

A line is drawn from the point $P\left(1,1 , 1\right)$ and perpendicular to a line whose direction ratios are $1,1,1$ to intersect the plane $x+2y+3z=4$ at the point $Q.$ The locus of $Q$ is

• A. $\frac{x}{1}=\frac{y - 5}{- 2}=\frac{z + 2}{1}$
• B. $\frac{x}{- 2}=\frac{y - 5}{1}=\frac{z + 2}{1}$
• C. $x=y=z$
• D. $\frac{x}{2}=\frac{y}{3}=\frac{2}{5}$

Answer 1

Answer: (A)

Let point $Q$ is $\left(x_{1} , y_{1} , z_{1}\right)$
D.R’s of $PQ$ are $x_{1}-1,y_{1}-1,z_{1}-1$
$\Rightarrow 1\left(x_{1} - 1\right)+1\left(y_{1} - 1\right)+1\left(z_{1} - 1\right)=0$
$\Rightarrow x_{1}+y_{1}+z_{1}=3$
Also $x_{1}+2y_{1}+3z_{1}=4$
$x_{1}+y_{1}+z_{1}=3$
$\Rightarrow \left(x_{1} , y_{1} , z_{1}\right)$ is a point on the two planes
$x+2y+3z=4$
$x+y+z=3$
$\therefore $ the line on the intersection of these two planes will be the required locus
Now, the direction vector of the line is $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix}=\hat{i}-2\hat{j}+\hat{k}$
Also, the point $\left(0 , 5 , - 2\right)$ satisfy both the equation of planes
Hence, the equation of the required line is $\frac{x}{1}=\frac{y - 5}{- 2}=\frac{z + 2}{1}$