Hi,
Your subject title is correct. Vectors are the way to solve the
problems. Here is the solution to problem 1.
1. Find an equation for the plane that passes through the point Q1 = (-1,2,1)
and contains the interesection of the the planes P1: x+y-z=2 and P2: 2x-y+3z=1.
First, find the line of intersection of the two planes. Any line in
the plane that passes thru the foot (tail) of the normal
(perpendicular) vector to the plane is perpendicular to the normal
vector. So the line of intersection will be normal to the normal
vectors of both planes. Finding the normal vectors is easy. They
have the same coefficients as the coefficients of the plane.
v1 = <1, 1, -1>
v2 = <2, -1, 3>
To get the vector normal to v1 and v2, take the cross product.
n = v1 cross v2 = 2i -5j -3k = <2, -5, -3>
n is the directional vector. Now we need to find a point on the line.
Let z = 0. Then,
x + y = 2
2x - y = 1
3x = 3
x = 1
y = 1
The point Q2 = (1, 1, 0) is on the line. You can check and see that
it is in both planes.
The equation of the line then is:
Line: <1, 1, 0> + tn = <1, 1, 0> + t<2, -5, -3>
Now we have two points in the plane.
Q1=(-1,2,1) and Q2=(1, 1, 0).
Vector q1q2 = <2, -1, -1>
Now we have two vectors in the plane.
n = <2, -5, -3>
q1q2 = <2, -1, -1>
If we take the cross product we get the normal vector to the required plane.
n cross q1q2 = <2, -4, 8>
Since we only care about direction, not magnitude, we can reduce by
dividing by two and get <1, -2, 4>.
Taking the normal vector to the plane and one point on the plane, the
equation of the plane is:
1(x - 1) -2(y - 1) + 4(z - 0) = 0
or
x - 2y + 4z + 1 = 0
Here I plugged in point Q2. If I had plugged in point Q1 I would have
gotten the same equation. See below.
1(x + 1) - 2(y - 2) + 4(z - 1) = 0
x - 2y + 4z + 1 = 0
Hope this helps. |