Find the vector equation of the plane that contains the points A(2,5,6) B(-7,1,4) and C(6,-2,-9).?

Question

Does it contain the line with equation
[x,y,z] = [-3,-6,-11]+k[22,1,-11]? varify.

Answer 1

Given points A(2,5,6); B(-7,1,4); and C(6,-2,-9), find the vector equation of the plane containing them.

Let's create two vectors, u and v, from the points.

u = AB = <-7-2, 1-5, 4-6> = <-9, -4, -2>
v = AC = <6-2, -2-5, -9-6> = <4, -7, -15>

The vector equation of the plane P is:

P = A + su + tv = <2, 5, 6> + s<-9, -4, -2> + t<4, -7, -15>
P = <2 - 9s + 4t, 5 - 4s - 7t, 6 - 2s - 15t>
where s and t are scalars ranging over the real numbers.
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Does the plane P contain the line L with the equation

r = <-3,-6,-11> + k<22,1,-11>
where k is a scalar ranging over the real numbers?

If the given point Q(-3, -6, -11) is in the plane and the vector <22,1,-11> lies in the plane, then the line is in the plane.

First let's check the point.

P = <2 - 9s + 4t, 5 - 4s - 7t, 6 - 2s - 15t>
Q(-3, -6, -11)

Set each variable of the plane equal to the corresponding value for the point.

x: 2 - 9s + 4t = -3
y: 5 - 4s - 7t = -6
z: 6 - 2s - 15t = -11

y - 2z: -7 + 23t = 16
23t = 23
t = 1

Plug back into z to solve for s.
z: 6 - 2s - 15t = -11
6 - 2s - 15*1 = -11
-2s = -11 - 6 + 15 = -2
s = 1

Now plug into x to check for consistency.
x: 2 - 9s + 4t = -3
x: 2 - 9 + 4 = -3
-3 = -3

This is true so the point Q(-3, -6, -11) lies in the plane.

Now let's check the vector.

P = <2 - 9s + 4t, 5 - 4s - 7t, 6 - 2s - 15t>
Q(-3, -6, -11)

We have three equations in three unknowns. Notice we are only setting the components of the vectors equal.

x: - 9s + 4t = 22k
y: - 4s - 7t = 1k
z: - 2s - 15t = -11k

y - 2z: 23t = 23k
t = k

Plug back into z to solve for s.

z: - 2s - 15t = -11k
z: - 2s - 15k = -11k
-2s = 4k
s = -2k

Plug into x for consistency check.

x: - 9s + 4t = 22k
x: - 9(-2k) + 4k = 22k
22k = 22k

The vector also lies in the plane.

The point and vector both lie in the plane so the line lies in the plane.

Answer 2

You mean this, don't you? r ⋅ [( c - a ) × ( b - a )] = | [( c - a ) × ( b - a )] |^2 *[It is hard to read if you use a period for the dot product. If you type in " &s dot; " with the space removed, you will get the dot product symbol] The answer is no. The cross product ( c - a ) × ( b - a ) is normal to the plane because both (c - a) and (b - a) lie in the plane. If r is in the plane, it is perpendicular to ( c - a ) × ( b - a ) and the dot product is equal to zero. r ⋅ [( c - a ) × ( b - a )] = 0 You can also rewrite the equation as r = ˅n * |( c - a ) × ( b - a )| where ˅n is the unit normal vector At which point it is clear that r cannot be in the plane therefore any such equation is not an equation of the plane.