Question

Find the angle between the following pair of lines:
(i) $\overrightarrow{r}=2\hat{i}-5\hat{j}+\hat{k}+\lambda (3\hat{i}-2\hat{j}+6\hat{k})$ and $\overrightarrow{r}=7\hat{i}-6\hat{k}+\mu (\hat{i}+2\hat{j}+2\hat{k})$

(ii) $\overrightarrow{r}=3\hat{i}+\hat{j}-2\hat{k}+\lambda (\hat{i}-\hat{j}-2\hat{k})$ and $\overrightarrow{r}=2\hat{i}-\hat{j}-56\hat{k}+\mu (\widehat{3i}-5\hat{j}-4\hat{k})$

Answer 1

(i)

If $\theta$ be the angle between the given lines.
Then $\cos \theta =\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}\right|$
where $\overrightarrow{b_1}=(3\hat{i}+2\hat{j}+6\hat{k})$ and $\overrightarrow{b_2}=(\hat{i}+2\hat{j}+2\hat{k})$
$\therefore$ $\left|\overrightarrow{b_1}\right|=\sqrt{3^2+2^2+6^2}=7$
$\left|\overrightarrow{b_2}\right|=\sqrt{1^2+2^2+2^2}=3$
$\overrightarrow{b_1}.\overrightarrow{b_2}=(3\hat{i}+2\hat{j}+6\hat{k}).(\hat{i}+2\hat{j}+2\hat{k})$
$=3\times 1+2\times 2+6\times 2=3+4+12=19$
$\Rightarrow$ $\cos \theta =\frac{19}{7\times 3}$
$\Rightarrow$ $\theta ={\cos }^{-1}\left(\frac{19}{21}\right)$

(ii)

The given lines are parallel to the vectors,

$\overrightarrow{b_1}=(\hat{i}-\hat{j}-2\hat{k})$ and $\overrightarrow{b_2}=(3\hat{i}-5\hat{j}-4\hat{k})$ respectively.
$\therefore$ $\left|\overrightarrow{b_1}\right|=\sqrt{{\left(1\right)}^2+{(-1)}^2+{(-2)}^2}=\sqrt{6}$

$\left|\overrightarrow{b_2}\right|=\sqrt{{\left(3\right)}^2+{(-5)}^2+{(-4)}^2}=5\sqrt{2}$
$\overrightarrow{b_1}.\overrightarrow{b_2}=(\hat{i}-\hat{j}-2\hat{k}).(3\hat{i}-5\hat{j}-4\hat{k})$

$1.3-1(-5)-2(-4)=3+5+8=16$

If $\theta$ is the angle between the given lines then,
$\cos \theta =\left|\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}\right|$
$\Rightarrow$ $\cos \theta =\frac{16}{\sqrt{6}.5\sqrt{2}}=\frac{16}{\sqrt{2}.\sqrt{3}.5\sqrt{2}}=\frac{16}{10\sqrt{3}}$

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

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