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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
Question
Find the vector equation of a line passing through the point with position vector $\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$ and parallel to the line joining the points with position vectors $\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$ and $2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}.$ Also, find the cartesian equivalent of this equation.
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Answer

We know that, equation of a line passing through $\vec{\text{a}}$ and parallel to vector $\vec{\text{b}}$ is,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\dots(1)$
Here, $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
and, $\vec{\text{b}}=$ line joining $\big(\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
$=\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big)$
$=2\hat{\text{i}}-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{j}}+2\hat{\text{k}}-4\hat{\text{k}}$
$=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
Equation of the line is
$\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)$
For cortesion form of equation put $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(1+\lambda)\hat{\text{i}}+(-2+2\lambda)\hat{\text{j}}+(-3-2\lambda)\hat{\text{k}}$
Equating coeffcients of $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$ so
$\text{x}=1+\lambda,\text{y}=-2+2\lambda,\text{z}=-3-2\lambda$
$\Rightarrow\frac{\text{x}-1}{1}=\lambda,\frac{\text{y}+2}{2}=\lambda,\frac{\text{z}+3}{-2}=\lambda$
So, $\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}+3}{-2}$

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