# When Does a Normal Line Intersect The Plane

A line that is normal to the plane $4x-2y+5z-9=0$ passes through the origin. At what point does this normal line intersect the plane?

That equation represents the magnitude of the plane and the direction vector of the perpendicular line $(4,-2,5)$. I'm not sure how to solve this. I also do not have the answer so I cannot confirm if any of my answers are correct. Could someone show me how to solve this, or at least give me some ideas?

Thanks to anyone that helps.

Transform the equation into Hessian Normal Form - which is to say, divide $ax+by+cz=d$ by $\sqrt{a^2 + b^2 + c^2}$ so that you have an equivalent plane except that $\mathbf {\hat v} = \langle \hat a, \hat b, \hat c \rangle$ is a unit vector. Now, $\hat d$ is the distance from the origin to the plane -- and also the length of the vector parallel to the normal that goes from the origin to the plane, so $\hat d \mathbf {\hat v}$ is the location of the required point.

you have a direction $(4,-2,5)$ and a point on the line $\mathbf 0$

An equation for the line is: $x = 4t, y = -2t, z = 5t$

We must find a point on this line such that $4x- 2y + 5z -9 = 0$

Substitute $(x,y,z)$ from the equations of the line into the equation of the plane.

$45t-9 = 0\\ t = \frac 9{45}$

Now take this value of $t$ to find $(x,y,z)$