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VITEEE 2018 Paper

Section: Mathematics

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Question : 107 of 125

Marks: +1, -0

What is the length of the projection of

\hat{3i} + \hat{4j} + \hat{5k}

on the xy-plane ?

3
5
7
9

Solution:

xy-plane is perpendicular to z - axis. Let the vector

a =

\hat{3i} + \hat{4j} + \hat{5k}

make angle 0 with z - axis, then it makes 90-θ with xy-plane. unit vector along z-axis is k.

So, cos θ =

\frac{\hat{a} \cdot \hat{k}}{|\vec{a}| \cdot |\vec{k}|}

=

\frac{(3i+4j+5k) \cdot k}{|3i+4j+5k|}

=

\frac{5}{5\sqrt{2}}

=

\frac{1}{\sqrt{2}}

⇒ θ =

\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}

×

\frac{1}{\sqrt{2}}

= 5

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