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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
By computing the shortest distance determine whether the following pairs of lines intersect or not:
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
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Answer

Given equations of lines are,
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_1=\big(\hat{\text{i}}-\hat{\text{j}}\big),\vec{\text{b}}_1=\big(2\hat{\text{i}}+\hat{\text{k}}\big)$
and, $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_2=\big(2\hat{\text{i}}-\hat{\text{j}}\big),\vec{\text{b}}_2=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
We know that, shortest distance between lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\lambda\vec{\text{b}}_2$ is given by
$\text{S.D.}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|\dots(1)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(2\hat{\text{i}}-\hat{\text{j}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)$
$=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{i}}+\hat{\text{j}}$
$=\hat{\text{i}}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&-1 \end{vmatrix}$
$=\hat{\text{i}}(0-1)-\hat{\text{j}}(-2-1)+\hat{\text{k}}(2-0)$
$=-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=(\hat{\text{i}})\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)$
$=(1)(-1)+(0)(3)+(0)(2)$
$=-1+0+0$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=-1$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-1)^2+(3)^2+(2)^2}$
$=\sqrt{1+9+4}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{14}$
So, shortest distance between the given lines using equation (1) is,
$\text{S.D.}=\Big|\frac{-1}{\sqrt{14}}\Big|$
$=\frac{1}{\sqrt{14}}\text{ units}$
$\text{S.D.}\neq0$
Since, shortest distance between lines is not zero, so lines are not intersecting.

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