Question

The distance of the point $(2,3,4)$ from the plane $\overrightarrow{r}\cdot (3\hat{i}-2\hat{j}+6\hat{k})=5$ is

• A. $\frac{18}{7}$
• B. $\frac{19}{7}$
• C. $\frac{17}{7}$
• D. $\frac{16}{7}$

Answer 1

The correct option is B $\frac{19}{7}$

The distance of a point with position vector $\overrightarrow{a}$ from the plane

$\overrightarrow{r}$

$\overrightarrow{n}=d$, where $\overrightarrow{n}$ is the normal to the plane is $\left|\frac{\overrightarrow{a}\cdot \overrightarrow{n}-d}{|\overrightarrow{n}|}\right|$

Given the point is $(2,3,4)$

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

The equation of plane is $\overrightarrow{r}(3\hat{i}-2\hat{j}+6\hat{k})=5$

Comparing with $\overrightarrow{r}\cdot \overrightarrow{n}=d$

$\overrightarrow{n}=3\hat{i}-2\hat{j}=6\hat{k}$ and $d=5$

Distance of point from plane $=\left|\frac{\overrightarrow{a}\cdot \overrightarrow{n}-d}{|\overrightarrow{n}|}\right|$

$=\left|\frac{(2\hat{i}+3\hat{j}+4\hat{k})(3\hat{i}-2\hat{j}+6\hat{k})-5}{\sqrt{9+4+36}}\right|$

$=\left|\frac{6-6+24-5}{\sqrt{49}}\right|=\left|\frac{19}{\sqrt{49}}\right|=\frac{19}{7}$.