The problem I am faced with is:

Determine the scalar, vector, and parametric equations of a plane that passes through the points S(3,1,2) T(0,4,-2) and U(1,-3,5).

My attempt at solution:

ST = (3,1,2)-(0,4,-2)
ST = (3,-3,4)
SU = (3,1,2)-(1,-3,5)
SU = (2,4,-3)
n = (3,-3,4)-(2,4,-3)
n = (1,-7,7)

Scalar Equation:
1x - 7y + 7z + D = 0
1(3) - 7(1) + 7(2) + D = 0
D = 10
1x -7y + 7z - 10 = 0

Vector Equation:
(x,y,z) = (3,1,2) + s(3,-3,4) + t(2,4,-3)

Parametric Equation:
x = 3 + 3s + 2t
y = 1 - 3s + 4t
z = 2 + 4s - 3t

Where am I going wrong with this solution? Thanks!

  • $\begingroup$ for the normal to the plane, you want the cross-product, not the difference $\endgroup$ – J. W. Tanner Aug 29 at 3:54

You forgot to use the cross product to find the normal vector to plane.

Correctly got $SU=<2,4,-3>$ and $ST=<3,-3,4>$. The cross product gives you the normal vector which is $(7,-17,-18)$.

Use the normal vector to plug in a point in $7(x-x_1)-17(y-y_1)-18(z-z_1)=0$

note: cross product is defined as the the determinant of a matrix with first row the $i$ $j$ $k$ vectors the second row your first vector and in the third row your third vector. Order does not matter as both ways give a vector parallel to each other which is perpendicular to your original vectors.


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