Find the vector, or point orthogonal to plane 5x+3y+2y=0 and plane 5x+3y+2z=38 from point (0,0,0)
I have been playing with things for 2 hours.
update Sorry, I don't come here often. So, I had a few ideas.
One, I was playing with the distance equation such that $$ \frac{|Ax+By+Cz+D|}{\sqrt{A^2+B^2+C^2}} = Distance$$ but they wanted an exact vector on the plane.
So I was assuming that I should find an orthogonal vector to that point. Now I know the Vectors $$ P(5,3,2) and P_0(0,0,0) $$ So all I would have to do is find an n vector such that $$ \vec{n} (dot) (P-P_0) = 0 $$ but I have no idea how to incorporate 38 into it. I am so confused. I am not sure why
A normal vector (vector orthogonal to the plane) of $5x+3y+2z=0$ and $5x+3y+2z=38$ is $(5,3,2)$.
So, a vector orthogonal to the planes that begins in $(0,0,0)$ is $(5,3,2)$.
That's all there is to it.