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)

5

Answer 2

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