Question

The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line whose direction cosines are proportional to $2,3,-5$ is

• A. $\frac{9}{7}$
• B. $\frac{11}{7}$
• C. $\frac{15}{7}$
• D. None of these

Answer 1

Answer: (D)

None of these

Normal distance of the plane point from the plane will be

$=\frac{1+2+3-5}{\sqrt{3}}$

$=\frac{1}{\sqrt{3}}$ ...(i)

Now angle between the normal vector of the plane and direction vector of the line is

$(2\overrightarrow{i}+3\overrightarrow{j}-5\overrightarrow{k}).(\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})=|2\overrightarrow{i}+3\overrightarrow{j}-5\overrightarrow{k}||\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}|\cos \alpha$
$2-3-5=\sqrt{38}.\sqrt{3}.\cos \alpha$
$-6=\sqrt{38}\cos \alpha$
$\frac{-6}{\sqrt{38}}=\cos \alpha$...(ii)

Hence, the required distance '$d$' is

$d|\cos \alpha |=\frac{1}{\sqrt{3}}$

$d.\left(\frac{6}{\sqrt{38}}\right)=\frac{1}{\sqrt{3}}$

$d=\frac{\sqrt{38}}{6\sqrt{3}}$