Question

What is the length of the projection of $3\hat{i}+4\hat{j}+5\hat{k}$ on the xy-plane?

• A. 3
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (B)

$xy$ - plane is perpendicular to $z$ - axis.
Let the vector $a=3i+4j+5k$ make angle $\theta$ with $z$ -axis,
then it makes $90-\theta$ with $xy$ -plane.
unit vector along z-axis is $k$.
So, $\cos \theta =\frac{\vec{a}\cdot \hat{k}}{|\vec{a}|\cdot |\hat{k}|}=\frac{(3i+4j+5k)\cdot k}{|3i+4j+5k|}$
$=\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =\frac{\pi }{4}$
Hence angle with $xy$ - plane $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$
projection of $\vec{a}$ on $xy$ plane $=|\vec{a}|\cdot \cos \frac{\pi }{4}$
$=5\sqrt{2}\times \frac{1}{\sqrt{2}}=5$