Question

Distance of point $P(-2,3,4)$ from the line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ measured parallel to the plane $8x+6y-9z+1=0$ is

• A. $\frac{\sqrt{301}}{2}$
• B. $\frac{\sqrt{401}}{2}$
• C. $\frac{\sqrt{307}}{2}$
• D. $\frac{\sqrt{207}}{2}$

Answer 1

The correct option is B $\frac{\sqrt{401}}{2}$

Consider,

$\begin{array}{c}L=\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3x+4}{5}\\ P=8x+6y-9z+1=0\end{array}$

Let a point on line $L$ be $x=3t-2;y=\frac{4t-3}{2};z=\frac{5t-4}{3}$

Position vector from given to this point will be $1$ the plane $R$

Hence,

$\begin{array}{c}5\left(3t-2+2\right)+6\left(\frac{4t-3}{2}-3\right)-9\left(\frac{5t-4}{3}-4\right)=0\\ t=-1\end{array}$

Point $=\left(-5,\frac{-7}{2},-3\right)$

Distance from

$\begin{array}{c}\left(-2,3,4\right)=\sqrt{3^2+{\left(\frac{13}{2}\right)}^2+7^2}=\frac{\sqrt{401}}{2}\\ Hence,\\ OptionBiscorrectanswer.\end{array}$