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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.

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2 Answers 2

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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.

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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 $(x,y,z)$

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