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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry3 Marks
Question
Write the angle between the lines $\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}-2}{1}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}}{2}=\frac{\text{z}-1}{3}.$
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Answer

We have
$\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}-2}{1}$
$\frac{\text{x}-1}{1}=\frac{\text{y}}{2}=\frac{\text{z}-1}{3}$
The given lines are parallel to the vectors $\vec{\text{b}}_1=7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)}{\sqrt{7^2+(-5)^2+1^2}\sqrt{1^2+2^2+3^2}}$
$=\frac{7-10+3}{\sqrt{49+25+1}\sqrt{1+4+9}}$
$=0$

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