Question

The angles which the vector $\overrightarrow{A}=3\hat{i}+6\hat{j}+2\hat{k}$ makes with the coordinate axes are

• A. $co{s}^{-1}\frac{3}{7},co{s}^{-1}\frac{6}{7}$ and $co{s}^{-1}\frac{2}{7}$
• B. $co{s}^{-1}\frac{4}{7},co{s}^{-1}\frac{5}{7}$ and $co{s}^{-1}\frac{3}{7}$
• C. $co{s}^{-1}\frac{3}{7},co{s}^{-1}\frac{4}{7}$ and $co{s}^{-1}\frac{1}{7}$
• D. None of these

Answer 1

The correct option is A $co{s}^{-1}\frac{3}{7},co{s}^{-1}\frac{6}{7}$ and $co{s}^{-1}\frac{2}{7}$

If$\overrightarrow{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$

then, the angles which it makes with coordinate axes are given by:

$\cos \theta =\frac{\text{magnitude of component along concerned direction}}{A}$

where A is the magnitude of $\overrightarrow{A}$ is $A=\sqrt{3^2+6^2+2^2}=7$

therefore, angle with x- axis $\alpha ={\cos }^{-1}\frac{3}{7}$

angle with y-axix $\beta ={\cos }^{-1}\frac{6}{7}$

angle with z-axis $\gamma ={\cos }^{-1}\frac{2}{7}$