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QUESTION

# How do you find the equation of the plane in xyz-space through the point p=(4, 5, 4) and perpendicular to the vector n=(-5, -3, -4)?

The answer is: 5x+3y+4z-51=0

Given a poiint P(x_p,y_p,z_p) and a vector vecv(a,b,c) perpendicular to the plane, the equation is:

a(x-x_p)+b(y-y_p)+c(z-z_p)=0

So:

-5(x-4)-3(y-5)-4(z-4)=0rArr

-5x+20-3y+15-4z+16=0rArr

5x+3y+4z-51=0

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