Vectors help finale!?

Question

If P is the plane parallel to the plane with equation
4X1 + 5X2 + 6X3 = 7 and passing through the point (3, 1, 2).

Then what is the equation of P?


If Q (1, 1, 4), R (2, 3, 5), and S (-1, 0, -6) are three points in the Cartesian space. What is the equation of the plane they determine?

Answer 1

if they are parrellel, than u know the coeffiecents are the same, just the constant is different. So 4x1+5x2+6x3=constant. Plug in the point, and solve for the constant.

or
<4,5,6>*(x1-3,x2-1,x3-2)=0
they should give u the same thing. if not use the bottom method but check to make sure i wrote the formula right

Answer 2

If P is the plane parallel to the plane with equation
4X1 + 5X2 + 6X3 = 7 and passing through the point (3, 1, 2).
Then what is the equation of P?

I assume you mean that the variables are X1, X2, and X3.

A parallel plane will have the same coefficients for the three variables. Only the constant will change. Plug in the given point.

4(X1 - 3) + 5(X2 - 1) + 6(X3 - 2) = 0
4X1 - 12 + 5X2 - 5 + 6X3 - 12 = 0
4X1 + 5X2 + 6X3 - 29 = 0
____________

If Q(1, 1, 4), R(2, 3, 5), and S(-1, 0, -6) are three points in the Cartesian space. What is the equation of the plane they determine?

Create two vectors, u and v, that lie in the plane.

u = QR = R - Q = <2-1, 3-1, 5-4> = <1, 2, 1>
v = QS = S - Q = <-1-1, 0-1, -6-4> = <-2, -1, -10>

The cross product of two vectors in the plane will give you the normal vector n, to the plane.

n = u X v = <-19, 8, 3>

With the normal vector n and one point we can write the equation of the plane. Let's choose Q(1, 1, 4).

-19(x - 1) + 8(y - 1) + 3(z - 4) = 0
-19x + 19 + 8y - 8 + 3z - 12 = 0
-19x + 8y + 3z - 1 = 0

Answer 3

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