Question

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

Answer 1

$\overrightarrow{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (\hat{i}+2\hat{j}+3\hat{k})({\overrightarrow{a}}_1+\lambda \overrightarrow{b_1})$

$\overrightarrow{r}=\hat{i}-\hat{j}+\hat{k}+\mu (\hat{i}+\hat{j}+\hat{k})({\overrightarrow{a}}_1+\mu \overrightarrow{b_1})$

$\cos \theta =\left|\frac{{\overrightarrow{b}}_1.{\overrightarrow{b}}_2}{\left|{\overrightarrow{b}}_1\right|\left|{\overrightarrow{b}}_2\right|}\right|$

$a_1=\hat{i}+\hat{j}-\hat{k}{a}_2=3\hat{i}-\hat{j}+\hat{k}$

$a_1=\hat{i}+2\hat{j}+3\hat{k}{b}_2=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{b_1}.\overrightarrow{b_2}=(\hat{i}+2\hat{j}+3\hat{k}).(\hat{i}+\hat{j}+\hat{k})$

$=(1\times 1)+(2\times 1)+(3\times 1)$

$=1+2+3\Rightarrow 6$

$|\overrightarrow{b_1}|=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$

$|\overrightarrow{b_2}|=\sqrt{1+1+1}=\sqrt{3}$

$\cos \theta =\left|\frac{{\overrightarrow{b}}_1.{\overrightarrow{b}}_2}{\left|{\overrightarrow{b}}_1\right|\left|{\overrightarrow{b}}_2\right|}\right|$

$=\frac{6}{\sqrt{14}\sqrt{3}}$

$=\frac{2\sqrt{3}}{\sqrt{14}}=2\sqrt{\frac{3}{14}}$

$\therefore \theta ={\cos }^{-1}\left(2\sqrt{\frac{3}{4}}\right)$ seqn angle