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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry2 Marks
Question
Find the angle between the lines $\vec{\text{r}}=\big(2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$ and $\vec{\text{r}}=7\hat{\text{i}}-6\hat{\text{k}}+\mu\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$
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Answer

Let $\theta$ be the angle between the given lines. The given lines are parallel to the vectors $\vec{\text{b}}_1=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}},$ respectively.
So, the angle $\theta$ between the given lines is given by
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)}{\sqrt{3^2+2^2+6^2}\sqrt{1^2+2^+2^2}}$
$=\frac{3\times1+2\times2+6\times2}{\sqrt{49}\sqrt{9}}$
$=\frac{19}{21}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{19}{21}\big)$
Thus, the angle between the given lines is $\cos^{-1}\big(\frac{19}{21}\big).$

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