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