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.
x − y + 3z = 4
x + y + 2z = 2
3x + y + 7z = 9
adding #1 and #2 I got 2x + 5z = 6
adding #1 and #3 I got 4x + 10z = 13
You should realize that we cannot solve this any further, since both x and z would drop out
so lets find the line of intersection of plane #1 and #2
from 2x + 5z = 6
let z=2, then x = -2 and from #2, y = 0
let z=4, then x = -7 and from #2 y = 1
so we have two points on the line of intersection of the first two planes, namely
(-2,0,2) and (-7,1,4) from which we can get a direction vector (5,-1,-2)
and parametric equations of that line could be
x = -2+5t
y = -t
z = 2-2t
subbing that into the third equation 3x + y + 7z = 0 we get
-6 + 15t - t + 14 - 14t = 9
8=9 which is a contradiction.
Had this been a TRUE statement then all 3 planes would have intersected in the same line, like the spine of an open book
but the statement was FALSE, so the 3 planes must intersect in 3 different but parallel lines,
to find the other two line equations follow my procedure above for the other two possible pairs of planes.
As a check they should all have the same direction numbers.
BTW, this question is the hardest of the possible types of intersection of 3 planes.
To determine the intersection, if any, of the three planes, we can start by finding the normal vectors of the planes.
The given equations of the planes are:
1) x - y + 3z = 4
2) x + y + 2z = 2
3) 3x + y + 7z = 9
The normal vector of a plane is the coefficients of x, y, and z in its equation.
For plane 1:
Normal vector 1 = (1, -1, 3)
For plane 2:
Normal vector 2 = (1, 1, 2)
For plane 3:
Normal vector 3 = (3, 1, 7)
Now, let's check if the normal vectors are linearly dependent, which will help us determine the geometric interpretation and solutions of the system.
To check if the normal vectors are dependent, we can put them into a matrix and row reduce it.
Let's construct the matrix with the normal vectors as rows:
N = [[1, -1, 3], [1, 1, 2], [3, 1, 7]]
Now let's row reduce the matrix N:
RREF(N) = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]
Since the RREF of N is the identity matrix, the three normal vectors are linearly independent. Hence, the planes intersect at a single point.
To find the coordinates of the point, we can solve the system of equations formed by setting the given equations of the planes equal to each other:
x - y + 3z = 4 (Equation 1)
x + y + 2z = 2 (Equation 2)
3x + y + 7z = 9 (Equation 3)
We can use any method to solve these equations. Here, we will use the method of substitution.
From Equation 1, we can express x in terms of y and z:
x = 4 + y - 3z
Substituting this value of x into Equations 2 and 3:
(4 + y - 3z) + y + 2z = 2
3(4 + y - 3z) + y + 7z = 9
Simplifying these equations:
y - z = -2 (Equation 4)
3y - 2z = 5 (Equation 5)
Now, we can solve Equations 4 and 5 simultaneously to find the values of y and z.
Multiplying Equation 4 by 3 and adding it to Equation 5:
3y - 3z + 3y - 2z = -6 + 5
6y - 5z = -1
Simplifying this equation:
6y - 5z = -1
To solve for y, we can isolate y:
6y = 5z - 1
y = (5z - 1)/6
Substituting this value of y into Equation 4:
(5z - 1)/6 - z = -2
Multiply through by 6 to get rid of the denominator:
5z - 1 - 6z = -12
-z - 1 = -12
-z = -12 + 1
-z = -11
z = 11
Substituting the value of z back into y = (5z - 1)/6:
y = (5(11) - 1)/6
y = (55 - 1)/6
y = 54/6
y = 9
Finally, substituting the values of y and z into Equation 1:
x - y + 3z = 4
x - 9 + 3(11) = 4
x - 9 + 33 = 4
x + 24 = 4
x = 4 - 24
x = -20
Therefore, the intersection point of the three planes is (-20, 9, 11).
To determine the intersection, if any, for the given group of three planes, we will use the method of normal vectors.
Step 1: Find the normal vectors of the three planes.
The normal vector of a plane is given by the coefficients of x, y, and z in the equation of the plane. So, for the given planes, the normal vectors are:
Plane 1: (1, -1, 3)
Plane 2: (1, 1, 2)
Plane 3: (3, 1, 7)
Step 2: Check if the normal vectors are linearly independent.
To determine if the normal vectors are linearly independent, we can create a matrix using the vectors as columns and find its determinant. If the determinant is non-zero, then the vectors are linearly independent.
Using the given normal vectors, the determinant of the matrix is:
| 1 1 3 |
|-1 1 1 |
| 3 2 7 |
Calculating the determinant, we get: -8.
Since the determinant is non-zero (-8 ≠ 0), the normal vectors are linearly independent.
Step 3: Determine the geometric interpretation and number of solutions.
Since the normal vectors are linearly independent, the planes do not contain a common line or intersect at a single point. Instead, they intersect at a single line. Geometrically, the three planes do not coincide or intersect at a point, but their intersection forms a line.
Step 4: Determine a vector equation of the line of intersection.
To find a vector equation of the line of intersection, we need a point on the line and a direction vector for the line. We can find a point by solving the system of equations formed by the three planes.
Solving the system of equations:
x − y + 3z = 4 (Equation 1)
x + y + 2z = 2 (Equation 2)
3x + y + 7z = 9 (Equation 3)
Multiplying Equation 2 by (-1) and adding it to Equation 1, we get:
-2y + z = 2 (Equation 4)
Multiplying Equation 2 by 3 and subtracting it from Equation 3, we get:
-2y + z = 3 (Equation 5)
From Equations 4 and 5, we notice that they represent the same line. This tells us that there is no unique solution for the system of equations, and the planes intersect along a line.
To find a point on the line of intersection, we can set z = 0 in Equation 4 (or Equation 5) and solve for y:
-2y + 0 = 2
-2y = 2
y = -1
Substituting the value of y in Equation 4, we can solve for x:
-2(-1) + z = 2
2 + z = 2
z = 0
Thus, we have found a point on the line of intersection: (x, y, z) = (2, -1, 0).
Now, we need a direction vector for the line. We can use the cross product of the normal vectors to obtain the direction vector. Taking the cross product of the normal vectors (1, -1, 3) and (1, 1, 2), we get:
(-5, -1, 2)
Therefore, a vector equation of the line of intersection is given by:
r = (2, -1, 0) + t(-5, -1, 2), where t is a parameter.
To summarize:
- The three planes intersect along a single line.
- A vector equation of the line is r = (2, -1, 0) + t(-5, -1, 2).
- There is no unique solution for the system of equations represented by the planes.