Physics - Vectors Question with Solution | TestHub

The given vector is $\vec{B}=3\hat{i}+2\hat{j}+4\hat{k}$

The magnitude of the vector is $|\vec{B}|=\sqrt{3^2+2^2+4^2}=\sqrt{9+4+16}=\sqrt{29}$

The unit vector along the Y-axis is $\hat{j}$

The angle $\theta$ made by the vector with the Y-axis is given by $\cos \theta =\frac{\vec{B}\cdot \hat{j}}{|\vec{B}||\hat{j}|}$

$\cos \theta =\frac{2}{\sqrt{29}}$

$\theta ={\cos }^{-1}\left(\frac{2}{\sqrt{29}}\right)$

Answer: ${\cos }^{-1}\left(\frac{2}{\sqrt{29}}\right)$