One way of describing a plane in R3 is to give a point on the plane and 2 non-pa

Question

One way of describing a plane in R3 is to give a point on the plane and 2 non-parallel vectors in the plane. If we specify the point (1, -2, 3) to be on the plane, and if vectors A and B are given as A vector = 2 ihat - 3 jhat + 6 khat and B vector = -ihat + 2 jhat + 3 khat and R vector = x ihat + y jhat + z khat, Rvector0 = ihat - 2 jhat + 3 khat, then the vector equation R vector - R vector0 = s A vector + t B vector describes a plane. The real numbers s and t are parameters, and we have the equations defining a plane parametrically x - 1 = 2s - t y + 2 - -3 s + 2 t z - 3 = 6 s + 3t Use the above description of the plane to find an equation in the form a x + b y + c z + d = 0, which defines the same plane.

Explanation / Answer

We should have:

a(2s - t) + b(-3s+2t) + c(6s+3t) = 0

(2a-3b+6c) = 0

(-a+2b+3c) = 0

Multiply the second equation by 2 and add the result to the first:

b+12c = 0 -> b = -12c , a = 2b+3c = -21c

So one solution is:

(-21 , -12 , 1)

Therefore we have:

-21(2s - t) - 12(-3s+2t) + (6s+3t) = 0

or

-21(x-1) - 12(y+2) + (z-3) = 0

-21x - 12y + z + 21 - 24 - 3 = 0

so the plane equation is:

-21x - 12y + z - 6 = 0

a = -21

b = -12

c = 1

d = -6

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