Question 1201854: from "Vector Equations of Planes" lesson.
A plane passes through the points P(-2,3,1), Q(-2,3,2) and R (1,0,1)
a. using vectors PQ and vectors PR as your direction vectors, write the vector equation of this plane
b. using vector QR, and any other direction vector, write a second vector equation for this plane.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
Answer: < x,y,z > = < -2,3,1 > + s*< 0,0,1 > + t*< 3,-3,0 >
where s and t are any real numbers
Explanation:
The < -2,3,1 > is the position vector of point P. You can replace this position vector with the coordinates of Q or R.
You can pick any point in the plane.
This start position can be somewhat analogous to the y intercept
y = mx+b has m = slope and b = y intercept
b = start position
m = tells us how to move = direction vector
That's one way we can connect the ideas of 2D graphs with a 3D one like this.
The < 0,0,1 > represents the coordinates of vector PQ. It starts at P and points to Q.
Subtract corresponding coordinates to determine this vector
vector PQ = Q - P = < -2,3,2 > - < -2,3,1 > = < 0,0,1 >
The < 3,-3,0 > refers to vector PR which is calculated in a similar fashion.
The template can be written as such
< x,y,z > = positionVector + s*DirectionVector1 + t*DirectionVector2
and more specifically as this template
< x,y,z > = P + s*VectorPQ + t*VectorPR
We need 2 direction vectors because the plane is 2 dimensional.
We have 2 degrees of freedom of where to go along the flat surface.
Think of an xy axis.
The two direction vectors cannot lie on the same line.
Otherwise, infinitely many planes will result.
How can we determine if two vectors are on the same line or not?
By solving for k in this vector equation
PQ = k*PR
< 0,0,1 > = k*< 3,-3,0 >
< 0,0,1 > = < 3k, -3k, 0 >
It should be fairly clear that there aren't any solutions for k.
The last entries of 1 and 0 don't match up no matter what k would be.
Therefore, there is no way to scale PR to get PQ, and vice versa.
Furthermore, vectors PQ and PR are not on the same line.
This allows us to use them as direction vectors of the plane.
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Part (b)
Answer: < x,y,z > = < -2,3,1 > + s*< 0,0,-1> + t*< 3,-3,-1 >
where s and t are any real numbers
Explanation:
Start with the answer from part (a)
Flip the signs of vector PQ to get vector QP, so we get < 0,0,-1> as another possible direction vector.
Replace < 3,-3,0 >, which was from vector PR, with < 3,-3,-1 > which is vector QR.
The calculation of vector QR is similar to what is shown in part (a) when we found vector PQ.
The position vector can stay the same.
Although as mentioned earlier, you can replace the position vector with the coordinates from Q or R.
I'll let you check if vectors QP and QR are on the same straight line or not.
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