Question

Find the angle between the following pairs of lines:
(i) $\frac{x+4}{3}=\frac{y-1}{5}=\frac{z+3}{4}\mathrm{and}\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$

(ii) $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{-3}\mathrm{and}\frac{x+3}{-1}=\frac{y-5}{8}=\frac{z-1}{4}$

(iii) $\frac{5-x}{-2}=\frac{y+3}{1}=\frac{1-z}{3}\mathrm{and}\frac{x}{3}=\frac{1-y}{-2}=\frac{z+5}{-1}$

(iv) $\frac{x-2}{3}=\frac{y+3}{-2},z=5\mathrm{and}\frac{x+1}{1}=\frac{2y-3}{3}=\frac{z-5}{2}$

(v) $\frac{x-5}{1}=\frac{2y+6}{-2}=\frac{z-3}{1}\mathrm{and}\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5}$

(vi) $\frac{-x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}\mathrm{and}\frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}$

Answer 1

(i) $\frac{x+4}{3}=\frac{y-1}{5}=\frac{z+3}{4}\mathrm{and}\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$

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

$$\begin{gathered} \overrightarrow{b_1}=3\hat{i}+5\hat{j}+4\hat{k} \\ \overrightarrow{b_2}=\hat{i}+\hat{j}+2\hat{k} \end{gathered}$$

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

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(3\hat{i}+5\hat{j}+4\hat{k}\right).\left(\hat{i}+\hat{j}+2\hat{k}\right)}{\sqrt{3^2+5^2+4^2}\sqrt{1^2+1^2+2^2}} \\ =\frac{3+5+8}{10\sqrt{3}} \\ =\frac{8}{5\sqrt{3}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{8}{5\sqrt{3}}\right) \end{gathered}$$


(ii) $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{-3}\mathrm{and}\frac{x+3}{-1}=\frac{y-5}{8}=\frac{z-1}{4}$

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

Now,
$$\begin{gathered} \overrightarrow{b_1}=2\hat{i}+3\hat{j}-3\hat{k} \\ \overrightarrow{b_2}=-\hat{i}+8\hat{j}+4\hat{k} \end{gathered}$$

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

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(2\hat{i}+3\hat{j}-3\hat{k}\right).\left(-\hat{i}+8\hat{j}+4\hat{k}\right)}{\sqrt{2^2+3^2+{\left(-3\right)}^2}\sqrt{{\left(-1\right)}^2+8^2+4^2}} \\ =\frac{-2+24-12}{9\sqrt{22}} \\ =\frac{10}{9\sqrt{22}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{10}{9\sqrt{22}}\right) \end{gathered}$$


(iii) $\frac{5-x}{-2}=\frac{y+3}{1}=\frac{1-z}{3}\mathrm{and}\frac{x}{3}=\frac{1-y}{-2}=\frac{z+5}{-1}$

The equations of the given lines can be re-written as

$\frac{x-5}{2}=\frac{y+3}{1}=\frac{z-1}{-3}\mathrm{and}\frac{x}{3}=\frac{y-1}{2}=\frac{z+5}{-1}$

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

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

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

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(2\hat{i}+\hat{j}-3\hat{k}\right).\left(3\hat{i}+2\hat{j}-\hat{k}\right)}{\sqrt{2^2+1^2+{\left(-3\right)}^2}\sqrt{3^2+2^2+{\left(-1\right)}^2}} \\ =\frac{6+2+3}{\sqrt{14}\sqrt{14}} \\ =\frac{11}{14} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{11}{14}\right) \end{gathered}$$


(iv) $\frac{x-2}{3}=\frac{y+3}{-2},z=5\mathrm{and}\frac{x+1}{1}=\frac{2y-3}{3}=\frac{z-5}{2}$

The equations of the given lines can be re-written as

$\frac{x-2}{3}=\frac{y+3}{-2}=\frac{z-5}{0}\mathrm{and}\frac{x+1}{1}=\frac{y-{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{3}{2}}}=\frac{z-5}{2}$

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

Now,
$$\begin{gathered} \overrightarrow{b_1}=3\hat{i}-2\hat{j}+0\hat{k} \\ \overrightarrow{b_2}=\hat{i}+\frac{3}{2}\hat{j}+2\hat{k} \end{gathered}$$

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

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(3\hat{i}-2\hat{j}+0\hat{k}\right).\left(\hat{i}+{\displaystyle \frac{3}{2}}\hat{j}+2\hat{k}\right)}{\sqrt{3^2+{\left(-2\right)}^2+0^2}\sqrt{1^2+{\left({\displaystyle \frac{3}{2}}\right)}^2+2^2}} \\ =\frac{3-3+0}{\sqrt{13}\sqrt{{\displaystyle \frac{29}{4}}}} \\ =0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$


(v ) $\frac{x-5}{1}=\frac{2y+6}{-2}=\frac{z-3}{1}\mathrm{and}\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5}$

The equations of the given lines can be re-written as

$\frac{x-5}{1}=\frac{y+3}{-1}=\frac{z-3}{1}\mathrm{and}\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5}$

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

Now,
$$\begin{gathered} \overrightarrow{b_1}=\hat{i}-\hat{j}+\hat{k} \\ \overrightarrow{b_2}=3\hat{i}+4\hat{j}+5\hat{k} \end{gathered}$$

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

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(\hat{i}-\hat{j}+\hat{k}\right).\left(3\hat{i}+4\hat{j}+5\hat{k}\right)}{\sqrt{1^2+{\left(-1\right)}^2+1^2}\sqrt{3^2+4^2+5^2}} \\ =\frac{3-4+5}{\sqrt{3}\sqrt{{\displaystyle 50}}} \\ =\frac{4}{5\sqrt{6}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{4}{5\sqrt{6}}\right) \end{gathered}$$

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.


(vi) $\frac{-x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3}\mathrm{and}\frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}$

The equations of the given lines can be re-written as

$\frac{x-2}{2}=\frac{y-1}{7}=\frac{z+3}{-3}\mathrm{and}\frac{x+2}{-1}=\frac{y-4}{2}=\frac{z-5}{4}$

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

Now,

$$\begin{gathered} \overrightarrow{b_1}=2\hat{i}+7\hat{j}-3\hat{k} \\ \overrightarrow{b_2}=-1\hat{i}+2\hat{j}+4\hat{k} \end{gathered}$$

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

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|} \\ =\frac{\left(2\hat{i}+7\hat{j}-3\hat{k}\right).\left(-1\hat{i}+2\hat{j}+4\hat{k}\right)}{\sqrt{2^2+7^2+{\left(-3\right)}^2}\sqrt{{\left(-1\right)}^2+2^2+4^2}} \\ =\frac{-2+14-12}{\sqrt{62}\sqrt{21}} \\ =0 \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \end{gathered}$$