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)#?

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Answer 1 · 2015-02-19T16:35:28

Hello,

If the repair is orthonormal, then the plane has #-5x-3y-4z=d# where #d# is a real number.

Because the plane contains #p=(x=4,y=5,z=4)#, you can write

#-5\times 4 -3\times 5 -4\times 4 = d#

So you find # d=-51#. Conclusion an (not "the" !!) equation of your plane is #-5x-3y-4z=-51#.

Another one, easier, is #5x+3y+4z=51#.

Answer 2 · 2015-02-19T17:14:23

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#