Question

A line passing through the point with position vector $2\hat{i}-3\hat{j}+4\hat{k}$ and is perpendicular to the plane $\overrightarrow{r}.(3\hat{i}+4\hat{j}-5\hat{k})=7$. Find the equation of the line in Cartesian and vector forms.

Answer 1

The vector equation of a line passing through a point with position vector $\overrightarrow{a}$ and parallel to $\overrightarrow{b}$ is

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$

Given the line passes through $2\hat{i}-3\hat{j}+4\hat{k}$

So, $\overrightarrow{a}=2\hat{i}-3\hat{j}+4\hat{k}$

Finding normal of plane $\overrightarrow{r}.(3\hat{i}+4\hat{j}-5\hat{k})=7$

Comparing with $\overrightarrow{r}.\overrightarrow{n}=d$ we have

$\overrightarrow{n}=3\hat{i}+4\hat{j}-5\hat{k}$

Since line is perpendicular to the plane, the line will be parallel to the normal of the plane.

$\therefore \overrightarrow{b}=\overrightarrow{n}=3\hat{i}+4\hat{j}-5\hat{k}$

Hence, $\overrightarrow{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda (3\hat{i}+4\hat{j}-5\hat{k})$

$\therefore$ Vector equation of line is $\overrightarrow{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda (3\hat{i}+4\hat{j}-5\hat{k})$

Cartesian form is $x\hat{i}+y\hat{j}+z\hat{k}=(2+\lambda )\hat{i}+(-3+4\lambda )\hat{j}+(4-5\lambda )\hat{k}$

$\Rightarrow x=2+\lambda ,y=-3+4\lambda ,z=4-5\lambda$

$\Rightarrow \lambda =x-2=\frac{y+3}{4}=\frac{z-4}{-5}$ Which is the required cartesian equation of the line.

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