Question

Find the distance of the plane $2x-3y+4z-6=0$ from the origin.

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{6}{\sqrt{29}}$
• C. $\frac{4}{\sqrt{17}}$
• D. none of these

Answer 1

The correct option is B $\frac{6}{\sqrt{29}}$

The equation of the given plane is $2x-3y+4z-6=0$. The vector equation of this plane is $\overrightarrow{r}(2\hat{i}-3\hat{j}+4\hat{k})=6$.

Its normal is $\overrightarrow{r}(\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k})=\frac{6}{\sqrt{29}}$

Hence, the distance of the plane from the origin is $\frac{6}{\sqrt{29}}$.