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Straight line in space — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsStraight line in space5 Marks
Question
Find the vector equation of a line which is parallel to the vector $2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and which passes through the point (5, -2, 4). Also, reduce it to cartesian from.
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Answer

We know that, vector equation of line passing through a fixed point $\vec{\text{a}}$ and paralel to vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}},$ where $\lambda$ is scalar Here, $\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{a}}=5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$ So, rquation of required line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ $\vec{\text{r}}=\big(5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)$ Put $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$ so $\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)=(5+2\lambda)\hat{\text{i}}+(-2-\lambda)\hat{\text{j}}+(4+3\lambda)\hat{\text{k}}$ Comparing the cofficients of $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$ so $\text{x}=5+2\lambda,\text{y}=-2-\lambda,\text{z}=4+3\lambda$ $\Rightarrow\frac{\text{x}-5}{2}=\lambda,\frac{\text{y}+2}{-0}=\lambda,\frac{\text{z}-4}{3}=\lambda$ Cortesian form of equation of the line is,$\frac{\text{x}-5}{5}=\frac{\text{y}+2}{-0}=\frac{\text{z}-4}{3}$

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