Three Dimensional Geometry — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry3 Marks
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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