Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a

Home » Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Solution:

The angle between the lines with direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ is given by
$\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right|$
It is given that
$\begin{array}{l} a_{1}=a, b_{1}=b, c_{1}=c \\ a_{2}=b-c, b_{2}=c-a, c_{2}=a-b \end{array}$
Let’s now substitute the values in the above eq. we obtain,
$\begin{array}{l} \cos \theta=\left|\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}}\right| \\ =0 \\ \operatorname{Cos} \theta=0 \end{array}$
Therefore, $\theta=90^{\circ}[$ Since, $\cos 90=0$ ]
As a result, angle between the given pair of lines is $90^{\circ}$ .