Use normal vectors to determine the intersection, if any, for each of the following groups of three planes. Give a geometric interpretation in each case and the number of solutions for the corresponding linear system of equations. If the planes intersect in a line, determine a vector equation of the line. If the planes intersect in a point, determine the coordinates of the point.
2x − 5y + 3z = −10
3x + 6y + 3z = −6
3x − 4y + z = −1
None of the normals are scalar multiples of each other, so none of your planes are parallel
If we multiply the third equation by 3 we get 3z in all three equations.
#2 - #1 gave me x+11y = 4
the new #3 - #2 gave me 6x - 18y = 3
solving those two equations in x and y gave me x = 5/4 and y = 1/4
subbing that back into one of the originals I got z = -15/4
So your 3 planes intersect in a point, namely (5/4,1/4,-15/4)
To determine the intersection of the three planes, we can use the normal vectors of each plane. The normal vector of a plane represents the direction perpendicular to the plane.
First, let's find the normal vectors of the given planes.
For the plane 2x - 5y + 3z = -10, the coefficients of x, y, and z are the components of the normal vector, so the normal vector is (2, -5, 3).
Similarly, for the plane 3x + 6y + 3z = -6, the normal vector is (3, 6, 3).
And for the plane 3x - 4y + z = -1, the normal vector is (3, -4, 1).
A geometric interpretation for the intersection of planes can be as follows:
- If the three planes intersect at a point, it means they have a single common point in space.
- If the three planes intersect along a line, it means they have infinite common points, all lying on the same line.
- If the three planes do not intersect at all, it means they are parallel to each other and have no common points.
Now, let's determine the intersection for these specific planes:
To find if these planes intersect at a point or along a line, we can check if their normal vectors are linearly independent. If they are, the planes intersect at a point, and if they are not, the planes intersect along a line.
To check for linear independence, we can form a matrix with the normal vectors as its rows:
| 2 -5 3 |
| 3 6 3 |
| 3 -4 1 |
Now, we can row reduce this matrix to determine the number of linearly independent rows.
R2 = R2 - (3/2)R1
R3 = R3 - (3/2)R1
| 2 -5 3 |
| 0 9 -9 |
| 0 7 -2 |
R3 = R3 - (7/9)R2
| 2 -5 3 |
| 0 9 -9 |
| 0 0 1 |
From the row reduced echelon form of the matrix, we can see that there are three linearly independent rows. Therefore, the planes intersect at a point.
To find the coordinates of the point of intersection, we can solve the system of equations formed by equating the equations of the planes:
2x - 5y + 3z = -10 (1)
3x + 6y + 3z = -6 (2)
3x - 4y + z = -1 (3)
Let's solve these equations.
From equation (3), we can express x in terms of y and z:
x = (1/3)(4y - z - 1)
Substituting this into equations (1) and (2):
2[(1/3)(4y - z - 1)] - 5y + 3z = -10
3[(1/3)(4y - z - 1)] + 6y + 3z = -6
Simplifying and solving the system of equations:
(8/3)y - (2/3)z = -4
(4/3)y + (8/3)z = -2
Multiply the second equation by 3:
4y + 8z = -6
Multiply the first equation by 2 and subtract it from the second equation:
4y + 8z - (8y - 2z) = -6 - (-8)
-4y + 10z = 2
9z = -8
z = -8/9
Substituting the value of z into the equation -4y + 10z = 2:
-4y + (10)(-8/9) = 2
-4y - (80/9) = 2
-4y = 2 + (80/9)
-4y = 18/9 + 80/9
-4y = 98/9
y = -98/36
y = -49/18
Finally, substituting the values of y and z back into the equation for x:
x = (1/3)(4y - z - 1)
x = (1/3)(4(-49/18) - (-8/9) - 1)
x = (1/3)(-196/18 + 8/9 - 1)
x = (1/3)(-196/18 + 16/18 - 18/18)
x = (1/3)(-198/18)
x = -198/54
x = -11/3
Therefore, the coordinates of the point of intersection for the given planes are (-11/3, -49/18, -8/9).
In conclusion, the three planes intersect at a point, and the coordinates of the point of intersection are (-11/3, -49/18, -8/9).