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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
Question
Find the angle between the lines $\vec{\text{r}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}+\lambda(2\hat{\text{i}}+{\hat{\text{j}}}+2\hat{\text{k}})$ and $\vec{\text{r}}=(2\hat{\text{i}}-5\hat{\text{k}})+\mu(6\hat{\text{i}}+3{\hat{\text{j}}}+2\hat{\text{k}}).$
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Answer

We have $\vec{\text{r}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}+\lambda(2\hat{\text{i}}+{\hat{\text{j}}}+2\hat{\text{k}})$
And $\vec{\text{r}}=(2\hat{\text{i}}-5\hat{\text{k}})+\mu(6\hat{\text{i}}+3{\hat{\text{j}}}+2\hat{\text{k}})$
where $\vec{\text{a}_1}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}},\ \vec{\text{b}_1}=(2\hat{\text{i}}+{\hat{\text{j}}}+2\hat{\text{k}})$
And $\vec{\text{a}_2}=2\hat{\text{i}}-5\hat{\text{k}},\ \vec{\text{b}_2}=6\hat{\text{i}}+3{\hat{\text{j}}}+2\hat{\text{k}}$
If $\theta$ is angle between the lines, then
$\cos\theta=\frac{|\vec{\text{b}_1}\cdot\vec{\text{b}_2}|}{|\vec{\text{b}_1|}\cdot|\vec{\text{b}_1}|}$
$=\frac{|(2\hat{\text{i}}+{\hat{\text{j}}}+2\hat{\text{k}})\cdot|(6\hat{\text{i}}+3{\hat{\text{j}}}+2\hat{\text{k}})|}{|2\hat{\text{i}}+{\hat{\text{j}}}+2\hat{\text{k}}||6\hat{\text{i}}+3{\hat{\text{j}}}+2\hat{\text{k}}|}$
$=\frac{|12+3+4|}{\sqrt{9}\sqrt{49}}=\frac{19}{21}$
$\theta=\cos^{-1}\frac{19}{21}$

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