Question

The length of the projection of the line segment joining the points $(1,0,-1)$ and $(-1,2,2)$ on the plane $x+3y-5z=6$, is equal to:

• A. $2$
• B. $\sqrt{\frac{271}{53}}$
• C. $\sqrt{\frac{472}{31}}$
• D. $\sqrt{\frac{474}{35}}$

Answer 1

The correct option is D $\sqrt{\frac{474}{35}}$

Let $d$ be the distance between $A(1,0,-1)$ and $B(-1,2,2)$.

$d=\sqrt{(1+1{)}^2+(0-2{)}^2+(-1-2{)}^2}$

$d=\sqrt{17}$

$\overline{AB}=-2\hat{i}+2\hat{j}+3\hat{k}$

The vector along the normal to the plane $x+3y-5z=6$ is given by,

$\overline{n}=\hat{i}+3\hat{j}-5\hat{k}$

Let $\theta$ be the angle between the normal to the plane and the vector along $AB$.

We have,
$\cos \theta =\frac{\overrightarrow{n}.\overrightarrow{AB}}{|\overline{n}|\times |\overrightarrow{AB}|}$

$\cos \theta =\frac{(-2)+6+(-15)}{\sqrt{17\times 35}}$

$\cos \theta =\frac{-11}{\sqrt{595}}$

Hence, $\sin \theta =\sqrt{\frac{474}{595}}$

The projection of $AB$ on the plane is given by,

$d\times \sin \theta$

$=\sqrt{17}\times \sqrt{\frac{474}{595}}$

$=\sqrt{\frac{474}{35}}$

Hence, option D is correct