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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)

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

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  • $\begingroup$ Another way to do the computation is by putting the plane's equation in normal form. $\endgroup$
    – hardmath
    Oct 26 '20 at 14:40
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    $\begingroup$ Why PQ = (1,7,0)? $\endgroup$
    – George C
    Oct 26 '20 at 14:40
  • $\begingroup$ I think I mixed up things, I am suppose to do coordinates of Q - coordinates of P right ? $\endgroup$ Oct 26 '20 at 14:49
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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 $\frac{22}{\sqrt{35}}$.

(Don't forget the absolute value.)

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