# True or False: If the foot of perpendicular drawn from the origin to a plane is (5,-3,-2), then the equation of plane is $\overrightarrow{r}.(5\hat i-3\hat j-2\hat k)=38.$

Toolbox:
• Vector equation of a plane normal to unit vector $\hat n$ and a distance d from the orgin is $\overrightarrow r. \hat n=d$
The foot of perpendicular drawn from the orgin to a plane is $(5,-3,-2)$
The normal vector is $\overrightarrow n=5 \hat i-3 \hat j-2 \hat k$
The equation of a plane is $\overrightarrow r. \hat n-d=0$
=>$\overrightarrow r. \hat n=d$
Where $\overrightarrow n$ is the normal vector to the plane
$d=\bigg(\sqrt {5^2+(-3)^2+(-2)^2}\bigg)^2$
$\quad=\bigg(\sqrt {25+9+4}\bigg)^2$
$\quad=38$
Hence the required equation is $\overrightarrow r.(5 \hat i- 3 \hat j-2 \hat k)=38$
Hence the statement is $true$