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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Find the distance between the lines $l_1$ and $l_2$ given by$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$ and, $\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}+\mu\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
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Answer

$\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{a}}_2=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
$\vec{\text{a}}_2-\vec{\text{a}}_1=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
$=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&-1\\2&3&6\end{vmatrix}$
$=\hat{\text{i}}(6+3)-\hat{\text{j}}(12+2)+\hat{\text{k}}(6-2)$
$=9\hat{\text{i}}-14\hat{\text{j}}+4\hat{\text{k}}$
Shortest distance between 2 lines
$=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}}{\big|\vec{\text{b}}\big|}\Bigg|$
$=\Bigg|\frac{9\hat{\text{i}}-14\hat{\text{j}}+4\hat{\text{k}}}{\big|\sqrt{2^2+3^2+6^2}\big|}\Bigg|$
$=\Bigg|\frac{9\hat{\text{i}}-14\hat{\text{j}}+4\hat{\text{k}}}{\sqrt{49}}\Bigg|$
$=\Bigg|\frac{\sqrt{9^2+(-14)^2+4^2}}{\sqrt{49}}\Bigg|$
$=\Big|\frac{\sqrt{293}}{\sqrt{49}}\Big|=\frac{\sqrt{293}}{7}\text{ units}$

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