The angle between two vectors given by 6\hat{i} + 6\hat{j} …

Physics Vectors Multiplication of Vectors Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: September 12, 2026

The angle between two vectors given by 6\hat{i} + 6\hat{j} - 3\hat{k} and 7\hat{i} + 4\hat{j} + 4\hat{k} is

\(\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\)

\(\cos^{-1}\left(\frac{\sqrt{3}}{5}\right)\)

\(\sin^{-1}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\)

\(\sin^{-1}\left(\frac{\sqrt{5}}{3}\right)\)

Text Solution

Verified by Experts

The correct answer is:

\(\cos \theta = \frac{\overline{A} \cdot \overline{B}}{AB} = \frac{42 + 24 - 12}{\sqrt{36 + 36 + 9} \sqrt{49 + 16 + 16}} = \frac{54}{81}\)

\(\cos \theta = \frac{54}{81} \therefore \sin \theta = \frac{\sqrt{5}}{9} \text{ or } \theta = \sin^{-1} \frac{\sqrt{5}}{9}\)

\(\Rightarrow \cos \theta = \frac{1}{\sqrt{2}} \therefore \theta = 45^\circ\)

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