Find the angle between the following pairs of lines:
(i) $\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k})$ and
$\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$
(ii) $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k})$ and
$\vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})$
(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.
$\therefore\left|\vec{b}_{1}\right|=\sqrt{(1)^{2}+(-1)^{2}+(-2)^{2}}=\sqrt{6}$
$\left|\vec{b}_{2}\right|=\sqrt{(3)^{2}+(-5)^{2}+(-4)^{2}}=\sqrt{50}=5 \sqrt{2}$
$\vec{b}_{1} \cdot \vec{b}_{2}=(\hat{i}-\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}-4 \hat{k})$
$=1 \cdot 3-1(-5)-2(-4)$
$=3+5+8$
$=16$
$\cos Q=\left|\frac{\vec{b}_{1} \cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}\right|$
$\Rightarrow \cos Q=\frac{16}{\sqrt{6} \cdot 5 \sqrt{2}}=\frac{16}{\sqrt{2} \cdot \sqrt{3} \cdot 5 \sqrt{2}}=\frac{16}{10 \sqrt{3}}$
$\Rightarrow \cos Q=\frac{8}{5 \sqrt{3}}$
$\Rightarrow Q=\cos ^{-1}\left(\frac{8}{5 \sqrt{3}}\right)$