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.