Find the angle between the lines whose direction ratios are propo… - Vidyadip

Question

Find the angle between the lines whose direction ratios are proportional to a, b, c and b - c, c - a, a - b.

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Answer

Given, that the direction ratios of lines are proportional to a, b, c and b - c, c - a, a - b.
Let, $\vec{\text{x}}$ and $\vec{\text{y}}$ be the vector parallel to these lines respectively, so
$\vec{\text{x}}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
And, $\vec{\text{y}}=(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}$
Let, $\theta$ be the angle between $\vec{\text{x}}$ and $\vec{\text{y}}$, so,
$\cos\theta=\frac{\vec{\text{x}}\times\vec{\text{y}}}{\big|\vec{\text{x}}\big|\big|\vec{\text{y}}\big|}$
$=\frac{(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})[(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}]}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2+(\text{a}-\text{b})^2}}}$
$=\frac{(\text{a})(\text{b}-\text{c})+\text{b}(\text{c}-\text{a})+\text{c}(\text{a}-\text{b})}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2\sqrt{\text{b}^2+\text{c}^2-2\text{bc}+\text{c}^2+\text{a}^2-2\text{ac}+\text{a}^2+\text{b}^2-2\text{ab}}}}$
$\cos\theta=\frac{\text{ab}-\text{ac}+\text{bc}-\text{ba}+\text{ca}-\text{bc}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
Angle between the lines $=\frac{\pi}{2}$

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