Question
Find the angle between lines $\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})$
$\text { and } \bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+\hat{k})$

Answer

Let $\bar{b}$ and $\bar{c}$ be vectors along given lines.
$
\therefore \quad \bar{b}=2 \hat{i}-2 \hat{j}+\hat{k} \text { and } \bar{c}=\hat{i}+2 \hat{j}+2 \hat{k}
$
Angle between lines is same as the angle between $\bar{b}$ and $\bar{c}$.
The angle between $\bar{b}$ and $\bar{c}$ is given by,
$
\begin{aligned}
& \cos \theta=\frac{\bar{b} \cdot \bar{c}}{|\bar{b}| \cdot|\bar{c}|}=\frac{(2 \hat{i}-2 \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})}{3 \times 3}=\frac{0}{9}=0 \\
& \therefore \cos \theta=0 \quad \therefore \theta=90^{\circ}
\end{aligned}
$
Lines are perpendicular to each other.

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