This is another question I can't figure out.
Given planes pie1 and pie2 defined as:
pie1: 2x + 5y - 7x + 2 = 0
pie2: x + y - z + 3 = 0
(b) Give the equation of the line of intersection of these two planes.
I only found the answer for (a) where they ask to find the angle between both planes.
Thank you!
To find the equation of the line of intersection between the two planes, you can follow these steps:
Step 1: Rewrite the equations of the planes in standard form:
pie1: -5x + 5y + 2 = 0
pie2: x + y - z + 3 = 0
Step 2: From the equations, we can extract the direction vector for each plane by taking the coefficients of x, y, and z. For pie1, the direction vector is (1, 1, 0), and for pie2, the direction vector is (1, 1, -1).
Step 3: To find the direction vector of the line of intersection, we can take the cross product of the direction vectors of the two planes. The direction vector of the line will be orthogonal (perpendicular) to both planes.
Calculate the cross product:
(1, 1, 0) × (1, 1, -1) = (1, -1, -2)
Step 4: Now we have the direction vector of the line of intersection, (1, -1, -2). To find a point on the line, we need to solve the equations of the planes simultaneously.
Substitute one of the equations into the other and solve for one variable:
-5x + 5y + 2 = x + y - z + 3
Combine like terms:
-6x + 4y + z = 1
We can choose a variable to be a parameter, let's say x = t.
Solving for y and z:
y = -1 + 2t
z = -4 - t
Step 5: Now we have a point on the line, (t, -1 + 2t, -4 - t), and the direction vector of the line, (1, -1, -2). We can use the point-direction form to write the equation of the line:
x = t
y = -1 + 2t
z = -4 - t
So, the equation of the line of intersection of the two planes pie1 and pie2 is:
(x, y, z) = (t, -1 + 2t, -4 - t)
This equation represents a parametric form of the line, where t is any real number.
To find the equation of the line of intersection between two planes, you need to find a direction vector for the line. This can be done by taking the cross product of the normal vectors of the two planes.
First, let's get the normal vectors of the two planes:
Plane pie1: 2x + 5y - 7z + 2 = 0
The coefficients of x, y, and z represent the components of the normal vector. So the normal vector of pie1 is [2, 5, -7].
Plane pie2: x + y - z + 3 = 0
Again, the coefficients of x, y, and z represent the components of the normal vector. So the normal vector of pie2 is [1, 1, -1].
Next, take the cross product of the two normal vectors to find the direction vector of the line. The cross product of two vectors can be found using the determinant method:
(direction vector) = [2, 5, -7] x [1, 1, -1]
= (5 * (-1) - (-7) * 1, -7 * 1 - 2 * (-1), 2 * 1 - 5 * 1)
= (-5 - (-7), -7 - (-2), 2 - 5)
= (-5 + 7, -7 + 2, 2 - 5)
= (2, -5, -3)
Now we have the direction vector of the line of intersection, which is (2, -5, -3).
To find the equation of the line, we also need a point on the line. Since the line lies on both planes, we can choose any point that satisfies both plane equations. To make it easier, we can set z = 0 and solve for x and y in each plane equation:
In pie1: 2x + 5y - 7z + 2 = 0
2x + 5y - 7(0) + 2 = 0
2x + 5y + 2 = 0
2x = -5y - 2
x = (-5y - 2) / 2
x = -2.5y - 1
In pie2: x + y - z + 3 = 0
(-2.5y - 1) + y - 0 + 3 = 0
-2.5y + y + 2 = 0
-1.5y = -2
y = -2 / -1.5
y = 4/3
So when z = 0, we have x = -2.5(4/3) - 1 = -3.333 and y = 4/3.
Therefore, we can choose the point (-3.333, 4/3, 0) on the line.
Finally, we can write the equation of the line using the point and direction vector:
r = (-3.333, 4/3, 0) + t(2, -5, -3)
where r represents any point on the line, t is a parameter, and (2, -5, -3) is the direction vector we found earlier.
So the equation of the line of intersection between pie1 and pie2 is:
x = -3.333 + 2t
y = 4/3 - 5t
z = -3t