Question

Let $L_{1}$ be the projection and $L_{2}$ be the image of $z$ -axis in the plane $3x-4y+z+1=0.$ The distance of the point $\left(1,2 , 3\right)$ from the plane containing the lines $L_{1}$ and $L_{2}$ is

• A. $2$ units
• B. $4$ units
• C. $5$ units
• D. $6$ units

Answer 1

Answer: (A)

The required plane must contain $z$ -axis and parallel to normal vector of the given plane
A parallel vector to $z$ -axis is $ < 0,0,1>$
Normal vector to the given plane is $ < 3,-4,1>$
A normal vector to the required plane $=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ 3 & -4 & 1 \end{vmatrix}$
$=4\hat{i}+3\hat{j}$
$\because $ Required plane contain $z$ axis, so it must pass through $\left(0,0 , 0\right)$
Equation of the required plane is $4\left(x - 0\right)+3\left(y - 0\right)=0\Rightarrow 4x+3y=0$
So, the perpendicular distance from $\left(1,2 , 3\right)$ to $4x+3y=0$ is $\left|\frac{4 + 6}{5}\right|=2$ units