Question

If $P = (0,1,0)$ and $Q = (0,0,1)$ then the projection of $PQ$ on the plane $x + y + z = 3$ is

A. $2$

B. $\sqrt 2$

C. $3$

D. $\sqrt 3$

Answer 1

Hint : Projection is the length on a base that is formed by a line due to perpendicular light falling on the given line. Here $P = (0,1,0)$ and $Q = (0,0,1)$ is given so find out the vector $\overrightarrow {PQ}$ and use the projection method of the vector using the dot-product.

Complete step-by-step answer :

$P = (0,1,0)$ and $Q = (0,0,1)$ asking about the projection of $PQ$ on the plane $x + y + z = 3$ .

Since position vector of $P = (0,1,0)$ and $Q = (0,0,1)$

Now, find $\overrightarrow {PQ} = \overrightarrow Q - \overrightarrow P$

$\therefore \overrightarrow {PQ} = (0 - 0)\widehat i + (0 - 1)\widehat j + (1 - 0)\widehat k$

$= - \widehat j + \widehat k$

So, now, find out the magnitude of vector $\overrightarrow {PQ}$

$\therefore \left| {\overrightarrow {PQ} } \right| = \sqrt {{{( - 1)}^2} + {{(1)}^2}} = \sqrt 2$

Now, equation of plane is given $x + y + z = 3$ so, write the

Normal vector $\overrightarrow n = \overrightarrow i + \overrightarrow j + \overrightarrow k$

Also find magnitude $\left| {\overrightarrow n } \right| = \sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}} = \sqrt 3$

Now, taking dot product of $\overrightarrow {PQ}$ with $\overrightarrow n$ , then

$\because \overrightarrow {PQ} \cdot \overrightarrow n = ( - \widehat j + \widehat k) \cdot (i + \widehat j + \widehat k)$

$\Rightarrow \left| {\overrightarrow {PQ} } \right|\left| {\overrightarrow n } \right|\cos \theta = - 1 + 1$

$\Rightarrow \sqrt 2 \sqrt 3 \cos \theta = 0$

$\Rightarrow \cos \theta = 0$

$\therefore \theta = \dfrac{\pi }{2}$

Now, finally calculate the projection of $\overrightarrow {PQ}$ on the plane $x + y + z = 3$

$\therefore$ Projection = $\left| {\overrightarrow {PQ} } \right|\times \cos (90 - \theta )$

$= \sqrt 2 \sin \theta (\because \theta = \dfrac{\pi }{2})$

$= \sqrt 2 \times 1$

$= \sqrt 2$ units.

Hence, the projection of $PQ$ on the plane $x + y + z = 3$ is $\sqrt 2$ units.

So, the correct answer is “Option B”.

Note : When we take dot product of $\overrightarrow {PQ}$ with normal vector $\overrightarrow n$ then the angle that is formed is with the normal vector $\widehat n$ but when we find with projection with plane then the value of angle of $\overrightarrow {PQ}$ with plane be $(90 - \theta )$ . So, a student should not be confused with these angles.