# Find the angle between the following pair of lines: $\frac{\large x-2}{\large 2} = \frac{\large y-1}{\large 5} = \frac{\large z + 3}{\large -3} \: and\: \frac{\large x+2}{\large -1} = \frac{\large y-4}{\large 8} = \frac{\large z - 5}{\large 4}$

Toolbox:
• Angle between the line is given by
• $\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Let $L_1$ be the line :$\large\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$
Direction ratios of the vector $b_1$ in $L_1$ is $(2,5,-3)$
Let this be $(a_1,b_1,c_1)$
Let $L_2$ be the line :$\large\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$
Direction ratios of the vector $b_2$ in $L_2$ is $(-1,8,4)$
Let this be $(a_2,b_2,c_2)$
we know $\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Now substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
$\cos \theta=\bigg|\large\frac{2 \times -1+5 \times 8+-3 \times 4}{\sqrt {2^2+5^2+(-3)^2}\sqrt {(-1)^2+8^2+4^2}}\bigg|$
$\cos \theta=\bigg|\large\frac{-2+40-12}{\sqrt {4+25+9}\sqrt {1+64+16}}\bigg|$
$\cos \theta=\bigg|\large\frac{26}{\sqrt {38}\sqrt {81}}\bigg|$
$\cos \theta=\large\frac{26}{9\sqrt {38}}$
$=>\theta=\cos ^{-1} \bigg(\large\frac{26}{9 \sqrt {38}}\bigg)$
answered Jun 5, 2013 by