Question

The distance of the plane $\vec{r}\cdot \left(\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}\right)=1$ from the origin is

• A. $1$
• B. $7$
• C. $\frac{1}{7}$
• D. None of these

Answer 1

Answer: (A)

$\vec{r}\cdot \left(\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}\right)=1$
Here $\vec{n}=\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}$
$\left|\vec{n}\right|=\sqrt{{\left(\frac{2}{7}\right)}^2+{\left(\frac{3}{7}\right)}^2+{\left(\frac{-6}{7}\right)}^2}$
$=\sqrt{\frac{49}{49}}=1$
So, $\vec{n}$ is a unit vector. Hence given eq. of plane is in normal from.
$\therefore$ Distance from origin $=1$