Find the angle between the lines whose direction ratios are a, b,… - Vidyadip

Question

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

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Answer

Direction ratios of one line are a, b, c
⇒ A vector along this line is $\vec{\text{b}_1}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
Direction ratios of second line are b - c, c - a, a - b
⇒ A vector along second line is $\vec{\text{b}_2}=(\text{b - c})\hat{\text{i}}+(\text{c - a})\hat{\text{j}}+(\text{a - b})\hat{\text{k}}$
Let $\theta$ be the angle between the two lines, then
$\cos\theta=\frac{\big|\vec{\text{b}_1}.\vec{\text{b}_2}\big|}{\big|\vec{\text{b}_1}\big|.\big|\vec{\text{b}_2}\big|}=\frac{\text{a}(\text{b - c})+\text{b}(\text{c - a})+\text{c}(\text{a - b})}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{(\text{b - c})^2+(\text{c - a})^2+(\text{a - b})^2}}$
$=\frac{\text{ab - ac}+\text{bc - ab}+\text{ac - bc}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{(\text{b - c})^2+(\text{c - a})^2+(\text{a - b})^2}}=0=\cos90^{\circ}$
$\Rightarrow\ \ \theta=90^{\circ}$

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