An ant is moving along the vector $\vec{l_1} : \hat{i} - 2\hat{j} + 3\hat{k}$. Few sugar crystals are kept along the vector $\vec{l_2} = 3\hat{i} - 2\hat{j} + \hat{k}$ which is inclined at an angle $\theta$ with vector $\vec{l_1}$. Then find the angle $\theta$. Also find the scalar projection of $\vec{l_1}$ on $\vec{l_2}$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\theta = \cos^{-1}(5/7)$; Projection $= \frac{10}{\sqrt{14}}$ ##
$\theta = \cos^{-1} \left( \frac{\vec{l_1} \cdot \vec{l_2}}{|\vec{l_1}| |\vec{l_2}|} \right)$
$= \cos^{-1} \left( \frac{(\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} + \hat{k})}{|\hat{i} - 2\hat{j} + 3\hat{k}| |3\hat{i} - 2\hat{j} + \hat{k}|} \right)$
$= \cos^{-1} \left( \frac{3+4+3}{\sqrt{1+4+9}\sqrt{9+4+1}} \right)$
$ = \cos^{-1} \left( \frac{10}{14} \right)$
$= \cos^{-1} \left( \frac{5}{7} \right)$
Scalar projection of $\vec{l_1}$ on $\vec{l_2} = \frac{\vec{l_1} \cdot \vec{l_2}}{|\vec{l_2}|}$
$= \frac{(\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} + \hat{k})}{|3\hat{i} - 2\hat{j} + \hat{k}|}$
$= \frac{3 + 4 + 3}{\sqrt{9 + 4 + 1}} = \frac{10}{\sqrt{14}}$