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Straight line in space — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsStraight line in space5 Marks
Question
Find the shortest distance between the following pairs of parallel lines whose equations are:$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
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Answer

The vector equation of the given lines are
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)-\mu\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
These two lines pass through the points having position vectors $\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{a}}_2=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$ and are parallel to the vector $\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
Now,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}-3\hat{\text{j}}-4\hat{\text{k}}\big)\times\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&-4\\1&-1&1 \end{vmatrix}$
$=-7\hat{\text{i}}-5\hat{\text{j}}+2\hat{\text{k}}$
$\Rightarrow\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|=\sqrt{(-7)^2+(-5)^2+2^2}$
$=\sqrt{49+25+4}$
$=\sqrt{78}$
The shortest distance between the two lines is given by
$\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|}{\big|\vec{\text{b}}\big|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$

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