Question

Find the angle between the following pairs of lines:
(i) $\overrightarrow{r}=\left(4\hat{i}-\hat{j}\right)+\mathrm{\lambda }\left(\hat{i}+2\hat{j}-2\hat{k}\right)\mathrm{and}\overrightarrow{r}=\hat{i}-\hat{j}+2\hat{k}-\mathrm{\mu }\left(2\hat{i}+4\hat{j}-4\hat{k}\right)$

(ii) $\overrightarrow{r}=\left(3\hat{i}+2\hat{j}-4\hat{k}\right)+\mathrm{\lambda }\left(\hat{i}+2\hat{j}+2\hat{k}\right)\mathrm{and}\overrightarrow{r}=\left(5\hat{j}-2\hat{k}\right)+\mathrm{\mu }\left(3\hat{i}+2\hat{j}+6\hat{k}\right)$

(iii) $\overrightarrow{r}=\mathrm{\lambda }\left(\hat{i}+\hat{j}+2\hat{k}\right)\mathrm{and}\overrightarrow{r}=2\hat{j}+\mathrm{\mu }\left\{\left(\sqrt{3}-1\right)\hat{i}-\left(\sqrt{3}+1\right)\hat{j}+4\hat{k}\right\}$

Answer 1

(i) $\overrightarrow{r}=\left(4\hat{i}-\hat{j}\right)+\mathrm{\lambda }\left(\hat{i}+2\hat{j}-2\hat{k}\right)\mathrm{and}\overrightarrow{r}=\hat{i}-\hat{j}+2\hat{k}-\mathrm{\mu }\left(2\hat{i}+4\hat{j}-4\hat{k}\right)$

Let $\overrightarrow{b_1}\mathrm{and}\overrightarrow{b_2}$ be vectors parallel to the given lines.

Now,
$$\begin{gathered} \overrightarrow{b_1}=\hat{i}+2\hat{j}-2\hat{k} \\ \overrightarrow{b_2}=2\hat{i}+4\hat{j}-4\hat{k} \end{gathered}$$

If $\theta$ is the angle between the given lines, then

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(\hat{i}+2\hat{j}-2\hat{k}\right).\left(2\hat{i}+4\hat{j}-4\hat{k}\right)}{\sqrt{1^2+2^2+{\left(-2\right)}^2}\sqrt{2^2+4^2+{\left(-4\right)}^2}} \\ =\frac{2+8+8}{3\times 6} \\ =1 \\ \Rightarrow \theta =0° \end{gathered}$$

(ii) $\overrightarrow{r}=\left(3\hat{i}+2\hat{j}-4\hat{k}\right)+\mathrm{\lambda }\left(\hat{i}+2\hat{j}+2\hat{k}\right)\mathrm{and}\overrightarrow{r}=\left(5\hat{j}-2\hat{k}\right)+\mathrm{\mu }\left(3\hat{i}+2\hat{j}+6\hat{k}\right)$

Let $\overrightarrow{b_1}\mathrm{and}\overrightarrow{b_2}$ be vectors parallel to the given lines.

Now,
$$\begin{gathered} \overrightarrow{b_1}=\hat{i}+2\hat{j}+2\hat{k} \\ \overrightarrow{b_2}=3\hat{i}+2\hat{j}+6\hat{k} \end{gathered}$$

If $\theta$ is the angle between the given lines, then

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(\hat{i}+2\hat{j}+2\hat{k}\right).\left(3\hat{i}+2\hat{j}+6\hat{k}\right)}{\sqrt{1^2+2^2+2^2}\sqrt{3^2+2^2+6^2}} \\ =\frac{3+4+12}{3\times 7} \\ =\frac{19}{21} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{19}{21}\right) \end{gathered}$$

(iii) $\overrightarrow{r}=\mathrm{\lambda }\left(\hat{i}+\hat{j}+2\hat{k}\right)\mathrm{and}\overrightarrow{r}=2\hat{j}+\mathrm{\mu }\left\{\left(\sqrt{3}-1\right)\hat{i}-\left(\sqrt{3}+1\right)\hat{j}+4\hat{k}\right\}$

Let $\overrightarrow{b_1}\mathrm{and}\overrightarrow{b_2}$ be vectors parallel to the given lines.

Now,
$$\begin{gathered} \overrightarrow{b_1}=\hat{i}+\hat{j}+2\hat{k} \\ \overrightarrow{b_2}=\left(\sqrt{3}-1\right)\hat{i}-\left(\sqrt{3}+1\right)\hat{j}+4\hat{k} \end{gathered}$$

If $\theta$ is the angle between the given lines, then

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(\hat{i}+\hat{j}+2\hat{k}\right).\left(\left(\sqrt{3}-1\right)\hat{i}-\left(\sqrt{3}+1\right)\hat{j}+4\hat{k}\right)}{\sqrt{1^2+1^2+2^2}\sqrt{{\left(\sqrt{3}-1\right)}^2+{\left(\sqrt{3}+1\right)}^2+4^2}} \\ =\frac{\left(\sqrt{3}-1\right)-\left(\sqrt{3}+1\right)+8}{\sqrt{6}\sqrt{24}} \\ =\frac{6}{12} \\ =\frac{1}{2} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{3} \end{gathered}$$