Question

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

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

Answer 1

The correct option is D $1$

$\overrightarrow{r}\cdot (\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k})=1$
Substituting $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
$(x\hat{i}+y\hat{j}+z\hat{k})\cdot (\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k})=1$
$\Rightarrow \frac{1}{7}[2x+3y-6z]=1$
$\Rightarrow 2x+3y-6z-7=0$

Distance of a plane $ax+by+cz+d=0$ from origin is given by
$\frac{\left|d\right|}{\sqrt{a^2+b^2+c^2}}=\frac{|-7|}{\sqrt{2^2+3^2+6^2}}$
$=\frac{7}{\sqrt{4+9+36}}$
$=\frac{7}{7}=1$
hence option (A) is correct