How do I find the point-normal form of the equation of the plane containing the point (-3,-4,3) and perpendicular to (4,1,-2)?

Question

I know to use $(r - p) \cdot n) = 0$ $(r-p)*n)=0$ but I keep getting $4 x + y - 2 z = - 10$ $4x+y-2z=-10$ and being told that's incorrect. Where am I going wrong? Have I just not simplified enough?

Thanks :)

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Answer 1 · 2017-10-02T02:17:56

The point-normal form of the equation of a plane is:

$n_{x} (x - x_{0}) + n_{y} (y - y_{0}) + n_{z} (z - z_{0}) = 0$ $n_x(x-x_0)+ n_y(y-y_0)+ n_z(z-z_0) = 0$

where $< n_{x}, n_{y}, n_{z} >$ $< n_x, n_y, n_z >$ is the given normal vector and $(x_{0}, y_{0}, z_{0})$ $(x_0,y_0,z_0)$ is the given point.

Given the normal vector $< 4, 1, - 2 >$ $<4,1,-2>$ and the point $(- 3, - 4, 3)$ $(-3, -4, 3)$ , the point-normal form is:

$4 (x - (- 3)) + (y - (- 4)) - 2 (z - 3) = 0$ $4(x-(-3))+ (y-(-4))-2(z-3) = 0$

The above equation is in the requested point-normal form and it seems that you are trying to write the equation in the scalar form:

Ax+By+Cz = D

This is not the point-normal form.