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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry3 Marks
Question
Write the value of $\lambda$ for which the lines $\frac{\text{x}-3}{-3}=\frac{\text{y}+2}{2\lambda}=\frac{\text{z}+4}{2}$ and $\frac{\text{x}+1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}+6}{-5}$ are perpendicular to each other.
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Answer

 We have
$\frac{\text{x}-3}{-3}=\frac{\text{y}+2}{2\lambda}=\frac{\text{z}+4}{2}$
$\frac{\text{x}+1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}+6}{-5}$
The given lines are parallel to vector $\vec{\text{b}}_1=-3\hat{\text{i}}+2\lambda\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}_2=3\lambda\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}.$
For $\vec{\text{b}}_1\perp\vec{\text{b}}_2,$ we must have
$\vec{\text{b}}_1.\vec{\text{b}}_2=0$
$\Rightarrow\big(-3\hat{\text{i}}+2\lambda\hat{\text{j}}+2\hat{\text{k}}\big).\big(3\lambda\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}\big)=0$
$\Rightarrow-7\lambda-10=0$
$\Rightarrow\lambda=-\frac{10}7{}$

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