Question
Find the vector equation of the line which is parallel to the vector $3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and which passes throught the point (1, -2, 3).
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Answer
Let $\vec{\text{a}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}$
So, vector equation of the line, which is parallel to the vector $\vec{\text{a}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and passes through the vector $\vec{\text{b}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}$ is $\vec{\text{r}}=\vec{\text{b}}+\lambda\vec{\text{a}}.$
$\therefore\vec{\text{r}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}+\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}{\hat{\text{j}}}+\text{z}\hat{\text{k}})-(\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}})=\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
$\Rightarrow(\text{x}-1)\hat{\text{i}}+(\text{y}+2)\hat{\text{j}}+(\text{z}-3)\hat{\text{k}}=\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
So, vector equation of the line, which is parallel to the vector $\vec{\text{a}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and passes through the vector $\vec{\text{b}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}$ is $\vec{\text{r}}=\vec{\text{b}}+\lambda\vec{\text{a}}.$
$\therefore\vec{\text{r}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}+\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}{\hat{\text{j}}}+\text{z}\hat{\text{k}})-(\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}})=\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
$\Rightarrow(\text{x}-1)\hat{\text{i}}+(\text{y}+2)\hat{\text{j}}+(\text{z}-3)\hat{\text{k}}=\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
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