Find the angle between the following pairs of lines:

Solution:

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by, $\cos Q=\left|\frac{\vec{b}_{1} \cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}\right|$

The given lines are parallel to the vectors, $\vec{b}_{1}=3 \hat{i}+2 \hat{j}+6 \hat{k}$ and $\vec{b}_{2}=\hat{i}+2 \hat{j}+2 \hat{k}$, respectively.

$\therefore\left|\vec{b}_{1}\right|=\sqrt{3^{2}+2^{2}+6^{2}}=7$

$\left|\vec{b}_{2}\right|=\sqrt{(1)^{2}+(2)^{2}+(2)^{2}}=3$

$\vec{b}_{1} \cdot \vec{b}_{2}=(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})$

$=3 \times 1+2 \times 2+6 \times 2$

$=3+4+12$

$=19$

$\Rightarrow \cos Q=\frac{19}{7 \times 3}$

$\Rightarrow Q=\cos ^{-1}\left(\frac{19}{21}\right)$

(ii) The given lines are parallel to the vectors, $\vec{b}_{1}=\hat{i}-\hat{j}-2 \hat{k}$ and $\vec{b}_{2}=3 \hat{i}-5 \hat{j}-4 \hat{k}$, respectively.

$\therefore\left|\vec{b}_{1}\right|=\sqrt{(1)^{2}+(-1)^{2}+(-2)^{2}}=\sqrt{6}$

$\left|\vec{b}_{2}\right|=\sqrt{(3)^{2}+(-5)^{2}+(-4)^{2}}=\sqrt{50}=5 \sqrt{2}$

$\vec{b}_{1} \cdot \vec{b}_{2}=(\hat{i}-\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}-4 \hat{k})$

$=1 \cdot 3-1(-5)-2(-4)$

$=3+5+8$

$=16$

$\cos Q=\left|\frac{\vec{b}_{1} \cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}\right|$

$\Rightarrow \cos Q=\frac{16}{\sqrt{6} \cdot 5 \sqrt{2}}=\frac{16}{\sqrt{2} \cdot \sqrt{3} \cdot 5 \sqrt{2}}=\frac{16}{10 \sqrt{3}}$

$\Rightarrow \cos Q=\frac{8}{5 \sqrt{3}}$

$\Rightarrow Q=\cos ^{-1}\left(\frac{8}{5 \sqrt{3}}\right)$