Browse Questions

Find the distance from the origin to the plane $\overrightarrow{r}=(\overrightarrow{2i}+\overrightarrow{j}+\overrightarrow{5j})=7$

Toolbox:
• Equation of the plane whose perpendicular distance from the origin is $P$ and $\overrightarrow n$ is the unit normal vector to the plane from the origin. $\overrightarrow r.\overrightarrow n=p$ (vector equation) $lx+my+nz=p$ where $\overrightarrow n=l\overrightarrow i+m\overrightarrow j+n\overrightarrow k$ If $\overrightarrow n = a\overrightarrow i+b\overrightarrow j+c\overrightarrow k$ is not the unit normal vector, then the equation is of the form $\overrightarrow r.\overrightarrow n=q$ where $p=\large\frac{q}{|\overrightarrow n|}$ The cartesian equation is $ax+by+cz=q$
Step 1
The equation of the plane $\overrightarrow r.(2\overrightarrow i-\overrightarrow j+5\overrightarrow k)=7$ is of the form $\overrightarrow r.\overrightarrow n=q$
$\overrightarrow n = 2\overrightarrow i - \overrightarrow j + 5 \overrightarrow k\: \: q=7$
Step 2
The perpendicular distance from the origin to the plane is $p=\large\frac{|q|}{|\overrightarrow n |}=\large\frac{7}{\sqrt{4+1+25}}=\large\frac{7}{\sqrt{30}}$ units.

edited Jun 25, 2013