#### Find the angle between the lines $\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda \left ( 2 \hat{i}+\hat {j}+2\hat{k} \right )$ and $\vec{r}=\left (2 \hat{i}-5\hat{k} \right )+\mu \left ( 6\har{i}+3\hat {j}+2\hat{k} \right )$

Given, lines:

$\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )$

$\vec{r}=\left (2\hat{i}-5\hat{k} \right )+\mu\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )$

We are instructed to find the angle between the lines.

The line $\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )$ is parallel to the vector

$2\hat{i}+\hat{j}+2\hat{k}$

Let

$\vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}$

Then, we can say the line $\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )$   is parallel to vector $\vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}$

Similarly, let $\vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k}$

Then, we can say is $\vec{r}=2\hat{j}-5\hat{k}+\mu \left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )$  parallel to the vector $\vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k}$

If we take θ as the angle between the lines, then cosine θ is:

$\cos \theta = \frac{\vec{b_{1}}\vec{b_{2}}}{\left |\vec{b_{1}} \right |\left |\vec{b_{2}} \right |}$

Substituting the values of $\vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}$  and $\vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k}$ in the above equation,

We get

$\cos \theta=\frac{\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )}{\left | 2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |}$

Here,

$\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=\left ( 2 \times 6 \right )+\left ( 1 \times 3 \right )+\left ( 2 \times 2 \right )$

$\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=12+3+4$

$\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=19...........(i)$

Also,

$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{2^{2}+1^{2}+2^{2}}\sqrt{6^{2}+3^{2}+2^{2}}$

$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{4+1+4}\sqrt{36+9+4}$

$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{9}\sqrt{49}$

$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=3 \times 7$

$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=21............(ii)$

Substituting the values of $\cos \theta$ in equation (i) and (ii), we get

$\cos \theta=\frac{19}{21}$

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

Therefore, the angle between the lines is $\cos^{-1}\left (\frac{19}{21} \right )$