Question

A vector $\overrightarrow{A}=2\hat{i}+3\hat{j}+6\hat{k}$ makes an angle of $\beta$ with positive direction of x-axis. $\beta$ is equal to:

• A. ${\tan }^{-1}\frac{2}{7}$
• B. ${\sin }^{-1}\frac{2}{7}$
• C. ${\cos }^{-1}\frac{2}{7}$
• D. ${\cos }^{-1}\frac{4}{7}$

Answer 1

Answer: (C)

${\cos }^{-1}\frac{2}{7}$

$\begin{array}{c}OA=\left(2,3,6\right)\\ OX=\left(1,0,0\right)\\ \cos \theta =\frac{OA.OX}{|OA||OX|}\\ =\frac{2\times 1}{\sqrt{2^2+3^2+6^2}\sqrt{1^2}}\\ =\frac{2}{\sqrt{49}}=\frac{2}{7}\\ \therefore \theta ={\cos }^{-1}\frac{2}{7}\\ Hence,optionCisthecorrectanswer.\end{array}$