Question

Find the angle between the following pair of lines

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

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

Answer 1

Here, the equation of given lines are
$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}and\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$
[$\because If\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ be a line then its dr's is $(a_1,b_1,c_1)$]
$\therefore$ Direction ratios of two lines are (2, 5, -3) and (-1, 8, 4).
Let $\theta$ be the acute angle between the given lines, then
$cos\theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\Rightarrow cos\theta =\frac{2\times (-1)+5\times 8+(-3)\times 4}{\sqrt{2^2+5^2+(-3{)}^2}\sqrt{(-1{)}^2+8^2+4^2}}=\frac{-2+40-12}{\sqrt{4+25+9}\sqrt{1+64+16}}=\frac{26}{\sqrt{38}\sqrt{81}}=\frac{26}{9\sqrt{38}}\Rightarrow \theta =co{s}^{-1}\left(\frac{26}{9\sqrt{38}}\right)$

Here, the equation of given lines are
$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}and\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$
$\therefore$ Direction ratios of two lines are (2, 2, 1) and (4, 1, 8).
Let $\theta$ be the acute angle between the given lines, then
$cos\theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\Rightarrow cos\theta =\frac{2\times 4+2\times 1+1\times 8}{\sqrt{2^2+2^2+1^2}\sqrt{4^2+1^2+8^2}}=\frac{8+2+8}{\sqrt{4+4+1}\sqrt{16+1+64}}=\frac{18}{\sqrt{9}\sqrt{81}}=\frac{18}{3\times 9}=\frac{2}{3}\Rightarrow \theta =co{s}^{-1}\left(\frac{2}{3}\right)$