Question:

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

Updated On: Feb 15, 2025
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The Correct Option is B

Solution and Explanation

xy-plane is perpendicular to z - axis. Let the vector
a=3i+4j+5k\vec{a} = 3i + 4j + 5k make angle θ\theta with z - axis, then it makes 90θ90-\theta with xy-plane, unit vector along z-axis is k.
So, cosθ=a,k^a,k^=(3i+4j+5k).k3i+4j+5kcos\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|}
=552=12θ=π4= \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow \quad\theta = \frac{\pi}{4}.
Hence angle with xy- plane π2π4=π4\frac{\pi }{2}-\frac{\pi }{4} = \frac{\pi }{4}
projection of a\vec{a} on xy plane =a.cosπ4= | \vec{a} | .cos \frac{\pi }{4}
=52×12=5.= 5\sqrt{2} \times \frac{1}{\sqrt{2}} = 5.
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