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AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Physics

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Question : 88 of 160

Marks: +1, -0

Find the angle between the vectors

\vec{A}=2\hat{i}+4\hat{j}+4\hat{k}

\vec{B}=4\hat{i}+2\hat{j}-4\hat{k}

.

0°
45°
60°
90°

Solution:

Given,

\vec{A}=2\hat{i}+4\hat{j}+4\hat{k}

and

\vec{B}=4\hat{i}+2\hat{j}-4\hat{k}

.

Let

\theta

be the angle between

A

and

B

.

By using dot product,

A \cdot B = AB \cos \theta

,

where, A and B are magnitudes of A and B respectively.

Now,

A=\sqrt{2^{2}+4^{2}+4^{2}}=6

B=\sqrt{4^{2}+2^{2}+(-4)^{2}}=6

Substituting the values, in dot product, we get

(2\hat{i}+4\hat{j}+4\hat{k}) \cdot (4\hat{i}+2\hat{j}-4\hat{k}) = (6)

(6)

\cos \theta

8+8-16=36 \cos \theta

cosθ = 0

⇒ θ = 90°

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