Question 1201853: this is from my 'Cartesian Equation of a Plane' lesson.
a. Determine the Cartesian equation of the plane that contains the three points A(-2,3,1), B(3,4,5) and C(1,1,0).
b. Determine the Cartesian equation of the plane containing the point P(-1,1,0) and perpendicular to the line between the points A(1,2,1) and B(3,-2,0).
thank you.
Found 2 solutions by Alan3354, math_tutor2020:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! a. Determine the Cartesian equation of the plane that contains the three points A(-2,3,1), B(3,4,5) and C(1,1,0).
-----
|-2 3 5|
D = | 3 4 5|
| 1 1 0|
D = 24
----
| 1 3 5|
a = | 1 4 5| = 7
| 1 1 0|
----------------
|-2 1 5|
b = | 3 1 5| = 17
| 1 1 0|
----------------
|-2 3 1|
c = | 3 4 1| = -13
| 1 1 1|
------
---> 7x + 17y -13z = 24
---
This is easily done with an Excel sheet.
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b. Determine the Cartesian equation of the plane containing the point P(-1,1,0) and perpendicular to the line between the points A(1,2,1) and B(3,-2,0).
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answers:
(a) 7x+17y-13z = 24
(b) 2x-4y-z = -6
Other answers are possible.
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Explanation for part (a)
We have these three points
A(-2,3,1)
B(3,4,5)
C(1,1,0)
Subtract the coordinates of B and A to find the vector that points from A to B.
B-A = < 3,4,5 > - < -2,3,1 >
B-A = < 3,4,5 > + < 2,-3,-1 >
B-A = < 3+2,4-3,5-1 >
B-A = < 5,1,4 >
This says to go from A to B, we do three things:- Move 5 units along the positive x axis.
- Move 1 unit along the positive y axis.
- Move 4 units along the positive z axis.
Therefore, vector AB is < 5,1,4 >
The order in vector naming is important. "Vector AB" means we start at A and point to B.
While "vector BA" means we start at B and point at A.
Repeat similar steps to find vector AC = < 3,-2,-1 >
We found that
vector AB = < 5,1,4 >
vector AC = < 3,-2,-1 >
Next, take the cross product of these two vectors.
This will construct a vector perpendicular to both AB and AC,
Think of this new vector as a vertical pole out of the flat horizontal ground. Vectors AB and AC are entirely on the flat ground.
The cross product of vectors AB and AC is < 7, 17, -13 > as discussed in this lesson here
https://www.algebra.com/algebra/homework/Vectors/cross-product.lesson
This represents the normal vector to the plane.
We can think of the normal vector being of the form < a,b,c >
In this case < a,b,c > = < 7,17,-13 >
The normal vector tells us how to tilt the plane.
One template equation for a plane is
a(x - p) + b(y - q) + c(z - r) = 0
where
a,b,c = coordinates of the normal vector we just found
p,q,r = coordinates of a point that is on the plane
We have 3 choices for what we pick for p,q,r
I'll go for (p,q,r) = (-2,3,1) which is the location of point A.
You could pick the coordinates of B or C.
So,
a(x-p) + b(y-q) + c(z - r) = 0
7(x-p) + 17(y-q) - 13(z - r) = 0 ... plugging in coordinates from normal vector
7(x-(-2)) + 17(y-3) - 13(z - 1) = 0 ... plugging in coordinates from point A in the plane
7(x+2) + 17(y-3) - 13(z - 1) = 0
7x+14 + 17y-51 - 13z + 13 = 0
7x+17y-13z + 14-51+13 = 0
7x+17y-13z - 24 = 0
7x+17y-13z = 24
That is one possible answer for part (a).
Other answers are possible because we could scale the equation up or down by some factor.
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Explanation for part (b)
Let's find vector AB
vector AB = B-A
vector AB = < 3,-2,0 > - < 1,2,1 >
vector AB = < 3-1,-2-2,0-1 >
vector AB = < 2,-4,-1 >
The direction vector of line AB is < 2,-4,-1 >
Because this line is perpendicular to the plane we want, vector AB is a normal vector of this plane.
normal vector = < a,b,c > = < 2,-4,-1 >
point on plane = (p,q,r) = (-1,1,0)
Equation of the plane in cartesian form
a(x-p) + b(y-q) + c(z - r) = 0
2(x - p) - 4(y-q) - 1(z - r) = 0 ... plugging in coordinates from normal vector
2(x - (-1)) - 4(y - 1) - 1(z - 0) = 0 ... plugging in coordinates from point
2(x+1) - 4(y-1) - z = 0
2x + 2 - 4y + 4 - z = 0
2x-4y-z+6 = 0
2x+4y-z = -6
Further Reading
https://www.whitman.edu/mathematics/calculus_online/section12.05.html
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