# Find the distance between a point and a plane

I have been trying to solve this exercice and I would like to know if my solution is correct or not ?
Point is $$P = (1, 7, 4)$$
Plane has an equation : $$5x + 3y + z = 8$$
I have taken a random point on the plane $$Q = (1, 1, 0)$$ $$\overrightarrow{PQ} = <0, -6, -4>$$ Normal to the plane is : $$\vec{N} = <5, 3, 1>$$ (by reading the coefficients of the plane's equation)

$$Distance = \left\lvert \vec{PQ}*\frac{\vec{N}}{\lvert N \rvert} \right\rvert = \frac{0-18-4}{\sqrt{5²+3²+1²}} = \frac{22}{\sqrt{35}}$$

• Another way to do the computation is by putting the plane's equation in normal form. Oct 26 '20 at 14:40
• Why PQ = (1,7,0)? Oct 26 '20 at 14:40
• I think I mixed up things, I am suppose to do coordinates of Q - coordinates of P right ? Oct 26 '20 at 14:49

Your reasoning is correct, however the calculation of the vector $$\vec{PQ}$$ is not correct. $$\vec{PQ} = Q - P = (0, -6, -4)$$. Using this vector you will end up with the fraction $$\frac{22}{\sqrt{35}}$$.