Consider the planes given by the equations
2y−2x−z=2
x−2y+3z=7
(a) Find a vector v parallel to the line of intersection of the planes.
(b)Find the equation of a plane through the origin which is perpendicular to the line of intersection of these two planes.
For part a) I just used the cross product of the vectors and got -8i-7j-2k
Then for part b) I used the vector I got for part a) and the point (0,0,0) to get -8x-7y-2z=0
Both answers are wrong and I don't know why.
To find a vector parallel to the line of intersection of the planes, you correctly used the method of taking the cross product of the normal vectors of the planes. However, it seems that there might have been a sign error in your calculations. Let's go through the process again to find the correct answer.
First, let's write the equations of the planes in standard form:
Plane 1: 2y - 2x - z = 2
Plane 2: x - 2y + 3z = 7
To find the normal vectors of the planes, we can extract the coefficients of x, y, and z from the equations:
Plane 1: normal vector v₁ = ⟨-2, 2, -1⟩
Plane 2: normal vector v₂ = ⟨1, -2, 3⟩
Now, to find a vector parallel to the line of intersection of the planes, we can take the cross product of the normal vectors:
v = v₁ × v₂ = ⟨(2)(3) - (-2)(-2), (-1)(3) - (2)(1), (-2)(-2) - (2)(1)⟩
= ⟨14, -5, -2⟩
So, a vector parallel to the line of intersection of the planes is v = ⟨14, -5, -2⟩.
Moving on to part b, to find the equation of a plane through the origin that is perpendicular to the line of intersection, we can use the vector v we found in part a and the point (0, 0, 0) on the plane.
The equation of the plane would be in the form of Ax + By + Cz = 0. We can substitute the values of the vector v into this equation to find A, B, and C:
Ax + By + Cz = (14)(0) + (-5)(0) + (-2)(0) = 0
Therefore, the equation of the plane is 0x + 0y + 0z = 0, which simplifies to 0 = 0.
So, the correct equation of a plane through the origin that is perpendicular to the line of intersection of the two planes is 0 = 0.
For part (a), using the cross product is the correct approach. However, the vector you obtained (-8i - 7j - 2k) seems to be incorrect. Let's go through the steps again to find the correct vector.
Given the equations of the planes:
2y − 2x − z = 2 ...(1)
x − 2y + 3z = 7 ...(2)
To find a vector parallel to the line of intersection of the planes, we can take the cross product of the normal vectors of the planes. The normal vectors can be obtained by taking the coefficients of x, y, and z in each equation.
For plane (1):
Normal vector = < -2, 2, -1 >
For plane (2):
Normal vector = < 1, -2, 3 >
Now, we can take the cross product of the normal vectors:
v = < -2, 2, -1 > × < 1, -2, 3 >
To calculate this cross product, we can use the formula:
v = < (2)(3) - (-2)(-2), (-2)(1) - (-2)(3), (1)(-2) - (2)(2) >
= < 2 - 4, -2 + 6, -2 - 4 >
= < -2, 4, -6 >
Therefore, a vector parallel to the line of intersection of the planes is v = < -2, 4, -6 >.
For part (b), to find a plane through the origin that is perpendicular to the line of intersection of the planes, we can use the vector we found in part (a). Let's call this vector v.
Since the plane passes through the origin (0, 0, 0), the equation of the plane can be given as:
-2x + 4y - 6z = 0
So, the correct equation of the plane through the origin which is perpendicular to the line of intersection of the two planes is -2x + 4y - 6z = 0.
Your method is correct, except you must have made an arithmetic error,
I got (-4, -5, -2) as the cross-product
a) which we can change to (4,5,2)
b) since we have a point (0,0,0)
the equation of the required plane is
4x + 5y + 2z = 0
Another way:
Add the two equations
-x + 2z = 9
x = 2z-9
let z = 0, ----> x = -9
sub into 2nd equation, y = -8
let z = 4, ---> x = -1
sub into 2nd equation, y = 2
So we have 2 points on the line of intersection,
(-9, -8, 0) and (-1, 2,4)
and a direction vector would be
(8, 10, 4)
or reduced to (4,5,2) as above