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 $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 0& 1\\ 3& -4& 1\end{array}\right|$
$=4\hat{i}+3\hat{j}$
$\because$ 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