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)*n)=0# but I keep getting #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#

where #< n_x, n_y, n_z ># is the given normal vector and #(x_0,y_0,z_0)# is the given point.

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

#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.