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 the vector $\vec{l}_1$. Then find the angle $\theta$. Also find the scalar projection of $\vec{l}_1$ on $\vec{l}_2$.

$\cos^{-1}(\frac{5}{7}), \frac{10}{\sqrt{14}}$

The correct answer is Option (1) → $\cos^{-1}(\frac{5}{7}), \frac{10}{\sqrt{14}}$ ##

$\theta = \cos^{-1} \left( \frac{\vec{l}_1 \cdot \vec{l}_2}{|\vec{l}_1| \cdot |\vec{l}_2|} \right)$

$= \cos^{-1} \left( \frac{(\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} + \hat{k})}{|(\hat{i} - 2\hat{j} + 3\hat{k})| \cdot |(3\hat{i} - 2\hat{j} + \hat{k})|} \right)$

$= \cos^{-1} \left( \frac{3 + 4 + 3}{\sqrt{1 + 4 + 9} \sqrt{9 + 4 + 1}} \right) = \cos^{-1} \left( \frac{10}{14} \right)$

$= \cos^{-1} \left( \frac{5}{7} \right)$

Scalar projection of $\vec{l}_1$ on $\vec{l}_2 = \frac{\vec{l}_1 \cdot \vec{l}_2}{|\vec{l}_2|}$

$= \frac{(\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} - 2\hat{j} + \hat{k})}{|(3\hat{i} - 2\hat{j} + \hat{k})|}$

$= \frac{3 + 4 + 3}{\sqrt{9 + 4 + 1}} = \frac{10}{\sqrt{14}}$