If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (–4, 3, –6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Solution:

It is known to us that the angle between the lines with direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ is given by
$\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right|$
So, a line passing through $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ has direction ratios $\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right),\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right),\left(\mathrm{z}_{1}-\mathrm{z}_{2}\right)$ The direction ratios of line joining the points $A(1,2,3)$ and $B(4,5,7)$
$\begin{array}{l} =(4-1),(5-2),(7-3) \\ =(3,3,4) \\ \therefore \mathrm{a}_{1}=3, \mathrm{~b}_{1}=3, \mathrm{c}_{1}=4 \end{array}$
Direction ratios of line joining the points $C(-4,3,-6)$ and $B(2,9,2)$
$\begin{array}{l} =(2-(-4)),(9-3),(2-(-6)) \\ =(6,6,8) \\ \therefore \mathrm{a}_{2}=6, \mathrm{~b}_{2}=6, \mathrm{c}_{2}=8 \end{array}$
Let’s now substitute the values in the above equation we get,
$\begin{array}{l} \cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right| \\ \cos \theta=\left|\frac{3 \times 6+3 \times 6+4 \times 8}{\sqrt{3^{2}+3^{2}+4^{2}} \sqrt{6^{2}+6^{2}+8^{2}}}\right| \end{array}$
$\begin{array}{l} =\left|\frac{18+18+32}{\sqrt{9+9+16} \sqrt{36+36+64}}\right| \\ =\left|\frac{68}{\sqrt{34} \sqrt{136}}\right| \\ =\left|\frac{68}{\sqrt{34} \sqrt{4 \times 34}}\right| \\ =\left|\frac{68}{34 \times 2}\right| \\ \cos \theta=1 \end{array}$
Therefore, $\theta=0^{\circ}[$ since, $\cos 0$ is 1 $]$
As a result, Angle between the lines $\mathrm{AB}$ and $\mathrm{CD}$ is $0^{\circ}$ .