Question

Angle between the pair of lines $\frac{x-2}{1}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+1}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$

• A. ${\cos }^{-1}\left(\frac{3}{\sqrt{35}}\right)$
• B. ${\cos }^{-1}\left(\frac{26}{9\sqrt{38}}\right)$
• C. ${\cos }^{-1}\left(\frac{4}{\sqrt{38}}\right)$
• D. ${\cos }^{-1}\left(\frac{2\sqrt{2}}{\sqrt{19}}\right)$

Answer 1

Answer: (A)

${\cos }^{-1}\left(\frac{3}{\sqrt{35}}\right)$

Let $a_1$ and $a_2$ be the vectors parallel to the pair of lines.

Given lines are

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

$\therefore \overrightarrow{a_1}=1\hat{i}+5\hat{j}-3\hat{k}$ and $\therefore \overrightarrow{a_2}=-1\hat{i}+8\hat{j}+4\hat{k}$

Then,

$\overrightarrow{a_1}.\overrightarrow{a_2}=(1\hat{i}+5\hat{j}-3\hat{k}).(-1\hat{i}+8\hat{j}+4\hat{k})$

$\overrightarrow{a_1}.\overrightarrow{a_2}=1\times (-1)+5\times 8+(-3)\times 4$

$\overrightarrow{a_1}.\overrightarrow{a_2}=-1+40-12$

$\overrightarrow{a_1}.\overrightarrow{a_2}=27$

Now, $\overrightarrow{\left|a_1\right|}=\sqrt{1^2+5^2+{(-3)}^2}$ and $\overrightarrow{\left|a_2\right|}=\sqrt{{(-1)}^2+8^2+4^2}$

$\overrightarrow{\left|a_1\right|}=\sqrt{1+25+9}$

$\overrightarrow{\left|a_1\right|}=\sqrt{35}$

$\overrightarrow{\left|a_2\right|}=\sqrt{1+64+16}$

$\overrightarrow{\left|a_2\right|}=\sqrt{81}$

$\overrightarrow{\left|a_2\right|}=9$

Let angle $\theta$ between the given pair of lines

$\cos \theta =\left|\frac{\overrightarrow{a_1}.\overrightarrow{a_2}}{\left|\overrightarrow{a_1}\right|.\left|\overrightarrow{a_2}\right|}\right|$

$\Rightarrow \cos \theta =\frac{27}{9\sqrt{35}}$

$\Rightarrow \cos \theta =\frac{3}{\sqrt{35}}$

$\Rightarrow \theta ={\cos }^{-1}\frac{3}{\sqrt{35}}$

Hence, it is complete solution.