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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry5 Marks
Question
By computing the shortest distance determine whether the following pairs of lines intersect or not:3
$\frac{\text{x}-5}{4}=\frac{\text{y}-7}{-5}=\frac{\text{z}+3}{-5}$ and $\frac{\text{x}-8}{7}=\frac{\text{y}-7}{1}=\frac{\text{z}-5}{3}$
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Answer

Given lines are ,
$\frac{\text{x}-5}{4}=\frac{\text{y}-7}{-5}=\frac{\text{z}+3}{-5}=\lambda$ (say)
$\Rightarrow\text{x}=4\lambda+5,\text{y}=-5\lambda+7,\text{z}=-5\lambda-3$
$\Rightarrow\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$=(4\lambda+5)\hat{\text{i}}+(-5\lambda+7)\hat{\text{j}}+(-5\lambda-3)\hat{\text{k}}$
$\vec{\text{r}}=\big(5\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(4\hat{\text{i}}-5\hat{\text{j}}-5\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_1=\big(5\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big),\vec{\text{b}}_1=\big(4\hat{\text{i}}-5\hat{\text{j}}-5\hat{\text{k}}\big)$
and, $\frac{\text{x}-8}{7}=\frac{\text{y}-7}{1}=\frac{\text{z}-5}{3}=\mu$ (say)
$\Rightarrow\text{x}=7\mu+8,\text{y}=\mu+7,3\mu+5$
$\Rightarrow\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$=(7\mu+8)\hat{\text{i}}+(\mu+7)\hat{\text{j}}+(3\mu+5)\hat{\text{k}}$
$\vec{\text{r}}=\big(8\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}\big)+\mu\big(7\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}=\big(8\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}\big),\vec{\text{b}}_2=\big(7\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)$
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
$\text{S.D.}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|\dots(1)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(8\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}\big)-\big(5\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big)$
$=8\hat{\text{i}}+7\hat{\text{j}}+5\hat{\text{k}}-5\hat{\text{i}}-7\hat{\text{j}}+3\hat{\text{k}}$
$=3\hat{\text{i}}+8\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\4&-5&-5\\7&1&3 \end{vmatrix}$
$=\hat{\text{i}}(-15+5)-\hat{\text{j}}(12+35)+\hat{\text{k}}(4+35)$
$=-10\hat{\text{i}}-47\hat{\text{j}}+39\hat{\text{k}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(3\hat{\text{i}}+8\hat{\text{k}}\big)\big(-10\hat{\text{i}}-47\hat{\text{j}}+39\hat{\text{k}}\big)$
$=(3)(-10)+(0)(-4)+(8)(39)$
$=-30+312$
$=282$
Using equation (1) to get the shortest distance between the given lines, so
$\text{S.D.}=\Bigg|\frac{282}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$\text{S.D.}\neq0$
Since, the shortest distance between given lines is not equal to zero, so Given lines are not intersecting.

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