Q.Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $2x+3y+6z=49$.

The cartesian plane $lx+my+nz=p$ has vector form $\vec r\cdot(l\hat i+m\hat j+n\hat k)=p$.

The plane $3x+4y-2z=5$ has normal direction ratios $(3,4,-2)$.

Its vector equation is simply obtained by replacing $x,y,z$ with $\vec r\cdot\hat i, \vec r\cdot\hat j, \vec r\cdot\hat k$:

$\vec r\cdot(3\hat i+4\hat j-2\hat k) = 5$

Divide the plane equation by the magnitude of its normal vector, arranging so the constant on the right is positive.

Given plane: $x+2y-3z-6=0$, i.e. $x+2y-3z=6$.

The normal form of a plane is $lx+my+nz=p$ where $(l,m,n)$ are direction cosines of the normal (so $l^2+m^2+n^2=1$) and $p\ge 0$ is the perpendicular distance from the origin.

The normal vector to the given plane is $(1,2,-3)$. Its magnitude:

$\sqrt{1^{2}+2^{2}+(-3)^{2}} = \sqrt{1+4+9} = \sqrt{14}$

Divide both sides of $x+2y-3z=6$ by $\sqrt{14}$:

$\frac{1}{\sqrt{14}}x + \frac{2}{\sqrt{14}}y - \frac{3}{\sqrt{14}}z = \frac{6}{\sqrt{14}}$

The right-hand side $\dfrac{6}{\sqrt{14}}>0$, so this is already in normal form. Rationalising, $p=\dfrac{6}{\sqrt{14}}=\dfrac{6\sqrt{14}}{14}=\dfrac{3\sqrt{14}}{7}$, and the direction cosines of the normal are $\left(\dfrac{1}{\sqrt{14}},\dfrac{2}{\sqrt{14}},\dfrac{-3}{\sqrt{14}}\right)$.