The question is: Find the equation of the plane passing through the point (2, 1, 3) and is normal to the plane 3x – 7y + 5z = -55.
I tried using the formula r.n = a.n
Doing so, I get:
(x,y,z).(3,-7,5) = (2,1,3).(3,-7,5)
Hence, the equation is, 3x – 7y + 5z = 14.
I'm not sure if this correct because the normal that I found for my equation is parallel to the plane but not perpendicular to it.
Best Answer
Let's denote by $\alpha:Ax+By+Cz=D$ a plane which solves the question.
The plane contains the point $P(2,1,3).$ So it is $2A+B+3C=D.$
Now, it is perpendicular to the plane $\pi:3x - 7y + 5z = -55.$ So, we have that $(A,B,C)\cdot (3,-7,5)=0.$ That is: $3A-7B+5C=0.$
We get $C=\frac{7B-3A}{5}$ and $D=2A+B+\frac{21B-9A}{5}.$
So the infinitely many solutions are given by
$$Ax+By+\frac{7B-3A}{5}z=2A+B+\frac{21B-9A}{5}$$ where $(A,B)\in \mathbb{R}^2, (A,B)\ne (0,0).$ In other words
$$5Ax+5By+(7B-3A)z=A+26B$$ where $(A,B)\in \mathbb{R}^2, (A,B)\ne (0,0).$
One particular solution ($A=B=1$) is $$5x+5y+4z=27.$$