What is the length of the projection of $3\hat{i}+4\hat{j}+5\hat{k}$ on the xy-plane ?
- 3
- 5
- 7
- 9
The Correct Option is B
Solution and Explanation
$\vec{a} = 3i + 4j + 5k$ 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 $= | \vec{a} | .cos \frac{\pi }{4}$
$= 5\sqrt{2} \times \frac{1}{\sqrt{2}} = 5.$
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Concepts Used:
Distance of a Point from a Plane
The shortest perpendicular distance from the point to the given plane is the distance between point and plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the particular point to the particular plane. Let's see the formula for the distance between point and plane.
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