Question

The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is

• A. $\frac{1}{7}$
• B. $7$
• C. $\frac{7}{5}$
• D. $1$

Answer 1

The correct option is D $1$

Let Equation of line $PQ$ through $P(1,-2,3)$ and parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is
$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=\lambda$ (say)
Let any point on this line
$Q(2\lambda +1,3\lambda -2,-6\lambda +3)$
$Q$ lies on plane. $x-y+z=5$
$(2\lambda +1)-(3\lambda -2)+(-6\lambda +3)=5$
$\Rightarrow -7\lambda =-1$
$\Rightarrow \lambda =\frac{1}{7}$
Now, distance
$PQ=\sqrt{(2\lambda {)}^2+(3\lambda {)}^2+(6\lambda {)}^2}$
$=\sqrt{(4\lambda {)}^2+(9\lambda {)}^2+(36\lambda {)}^2}$
$=7\lambda =1$ unit