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
I did this question for you above at 9:57 pm
To determine the intersection, if any, for the given group of three planes, we can use normal vectors.
Step 1: Find the normal vectors of each plane.
The normal vector of a plane is a vector that is perpendicular to that plane. The coefficients of x, y, and z in the equation of the plane represent its normal vector. Therefore, we can extract the normal vectors for each plane.
For the first plane: 2x − 5y + 3z = −10
The normal vector is (2, -5, 3)
For the second plane: 3x + 6y + 3z = −6
The normal vector is (3, 6, 3)
For the third plane: 3x − 4y + z = −1
The normal vector is (3, -4, 1)
Step 2: Determine the intersection.
To determine the intersection, we need to check if the normal vectors of any two planes are parallel or not. If the normal vectors are not parallel, it means that the three planes intersect at a single point. If they are parallel, it means they either intersect along a line or are the same plane.
To check if two normal vectors are parallel, we can calculate the cross product of the two normal vectors. If the cross product is the zero vector, it means they are parallel.
Let's check each pair of normal vectors:
Cross product of the first and second normal vectors:
(2, -5, 3) × (3, 6, 3) = (39, -3, 27)
Cross product of the first and third normal vectors:
(2, -5, 3) × (3, -4, 1) = (7, 9, 17)
Cross product of the second and third normal vectors:
(3, 6, 3) × (3, -4, 1) = (-21, 6, -30)
None of the cross products are the zero vector, which means the normal vectors are not parallel.
Since the normal vectors are not parallel, the three planes intersect at a single point.
Step 3: Determine the coordinates of the point of intersection.
To find the point of intersection, we need to solve the system of equations formed by the three planes. We can use various methods such as substitution or elimination to solve the system of equations.
Substituting the values of x, y, and z from any one of the three given equations into the other two equations will yield the same result. Let's choose the first equation:
From the first equation: 2x - 5y + 3z = -10
Solving for x:
2x = 5y - 3z - 10
x = (5y - 3z - 10) / 2
Now, substitute the value of x into the other two equations:
Substituting x into the second equation:
3((5y - 3z - 10) / 2) + 6y - 3z = -6
Simplifying the equation:
15y - 9z - 30 + 6y - 3z = -6
21y - 12z = 24
Substituting x into the third equation:
3((5y - 3z - 10) / 2) - 4y + z = -1
Simplifying the equation:
15y - 9z - 30 - 8y + 2z = -2
7y - 7z = 28
Now we have a system of two equations with two variables:
21y - 12z = 24
7y - 7z = 28
We can solve this system of equations using any preferred method, such as substitution or elimination.
Elimination method:
Multiply the second equation by 3:
21y - 21z = 84
Now, subtract the second equation from the first:
(21y - 12z) - (21y - 21z) = 24 - 84
-12z + 21z = -60
9z = -60
z = -60 / 9
z = -20/3
Substituting the value of z back into the second equation:
7y - 7(-20/3) = 28
7y + 140/3 = 28
7y = 28 - 140/3
7y = (84 - 140) / 3
7y = -56/3
y = -8/3
Substituting the values of y and z into the first equation:
2x - 5(-8/3) + 3(-20/3) = -10
2x + 40/3 - 60/3 = -10
2x = -10 + 20/3 - 120/3
2x = (-30 + 20 - 120)/3
2x = -130/3
x = -65/3
Therefore, the coordinates of the point of intersection for the given group of three planes are:
x = -65/3, y = -8/3, z = -20/3
The geometric interpretation in this case is that the three planes intersect at a single point, which is (-65/3, -8/3, -20/3). This point represents the common solution to the system of equations formed by the three planes.