If $P=(0,1,0), Q=(0,0,1)$, then the projection of $P Q$ on the plane $x+y+z=3$ is
If $P=(0,1,0), Q=(0,0,1)$, then the projection of $P Q$ on the plane $x+y+z=3$ is
$2$
$\sqrt{2}$
$3$
$\sqrt{3}$
Solution
Direction ratios of $P Q$
$=(0-0,0-1,1-0)=(0,-1,1)$
Direction cosine $=\left(|0|,\left|-\frac{1}{\sqrt{2}}\right|,\left|\frac{1}{\sqrt{2}}\right|\right)$
The given plane is $x+y+z=3$
Direction ratios of the plane are $(1,1,1)$ Length of the projection
$\begin{aligned}
& =0.1+1 \cdot \frac{1}{\sqrt{2}}+1 \cdot \frac{1}{\sqrt{2}} \\
& =\frac{2}{\sqrt{2}}=\sqrt{2}
\end{aligned}$