Question

The angle subtended by the vector $\overline{A}=4\hat{i}+3\hat{j}+12\hat{k}$ with the x-axis is

• A. $si{n}^{-1}(\frac{3}{13})$
• B. $si{n}^{-1}(\frac{4}{13})$
• C. $co{s}^{-1}(\frac{4}{13})$
• D. $co{s}^{-1}(\frac{3}{13})$

Answer 1

The correct option is C $co{s}^{-1}(\frac{4}{13})$

Given,

$\overrightarrow{A}=4\hat{i}+3\hat{j}+12\hat{k}$

$\overrightarrow{B}=\hat{i}$

Dot product of the vector is given by

$\overrightarrow{A}.\overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|cos\theta$

$cos\theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}||\overrightarrow{B}|}$

$cos\theta =\frac{(4\hat{i}+3\hat{j}+12\hat{k}).(\hat{i}+0\hat{j}+0\hat{k})}{\sqrt{4^2+3^2+{12}^2}.\sqrt{1^2+0^2+0^2}}=\frac{4\times 1+3\times 0+12\times 0}{\sqrt{169}}$

$cos\theta =\frac{4}{13}$

$\theta =co{s}^{-1}\left(\frac{4}{13}\right)$

The correct option is C.