FP3 - perpendicular distance from a point to a plane

Hi,

I'm asked to find the perpendicular distance from the point $P(3,5,2)$ to the plane with equation $3x - 2y + z = 4$

Okay, so I said let Q be the point on the plane closest to P. This implies that $PQ$ will be perpendicular to the plane. I have the direction of this vector from my equation of the plane and I know a point where the line $PQ$ crosses as I know the coordinates of $P$. I can therefore say that the equation of the line $PQ$ is:



I can therefore find the coordinates of Q by seeing where this lines intersects the equation of the plane, i.e.:



Now, this is where the worked solution and I differ. What I thought you should do is plug the value of $t$ in the equation of the line for $PQ$ i.e.



However, what the worked solution does is it plugs this value of $t$ into the vector equation of $\vec{PQ}$, i.e.



I'm having difficulty understanding the difference between what these two equations are finding. What is my equation actually finding? Why does theirs work?

Thank you

The distance you want to find is |PQ|.

You have $\vec{OP}=\begin{pmatrix} 3 \\ 5 \\ 2 \end{pmatrix}$ and $\vec{OQ}=\begin{pmatrix} 3 \\ 5 \\ 2 \end{pmatrix}+\frac{3}{14}\begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}$.

So what is $\vec{PQ}$ and then what is |PQ|?

Ahhh! Thank you



Once I drew a diagram with the origin, $\mathbf{q}$ and $\mathbf{p}$ I could see what was happening.

Thanks again