Find two vectors perpendicular to both (3, -6, 3) and (-2, 4, 2).
argghhhh! wrong again.
first reduce you starting vectors to simplest form of (1, -2, 1) and (1, -2,-1)
a vector perpendicular to these two is (4, 2, 0)
which reduces to (2, 1, 0)
so any scalar multiple of would work , such as (6, 3, 0) or (-2, -1, 0) etc
give it up , bot
but you had an answer of (-4, 0, 0)
which DOES NOT reduce to my answers
So you are wrong again.
I apologize for the repeated errors in my responses. Thank you for your correction.
Let's first find the cross product of the two given vectors:
(3, -6, 3) x (-2, 4, 2) =
[(−6)(2) − (3)(4), (3)(−2) − (3)(−2), (3)(4) − (−6)(−2)] =
(-12, 0, 18)
Now we need to find two vectors that are perpendicular to (-12, 0, 18). One way to do this is to choose two of the coordinate planes (xy, xz, or yz) and set one coordinate to zero while solving for the remaining two coordinates.
Here are two possible vectors:
- (1, 0, 0): This vector lies entirely in the xz-plane and is perpendicular to the y-axis.
- (0, -3, 2): This vector lies in both the xy- and yz-planes and is perpendicular to the x-axis.
Both of these vectors are perpendicular to both (3, -6, 3) and (-2, 4, 2).
I'm sorry, you are correct! Thank you for correcting me.
To find two vectors perpendicular to both (1, -2, 1) and (1, -2, -1), we first reduced the given vectors to their simplest form. Then we found a vector that is perpendicular to both of these vectors by taking their cross product:
(1, -2, 1) x (1, -2, -1) = (-4, 0, 0)
Now we need to find two vectors that are perpendicular to (-4, 0, 0). One way to do this is again to choose two of the coordinate planes (xy, xz, or yz) and set one coordinate to zero while solving for the remaining two coordinates.
Here are two possible vectors:
- (0, 1, 2): This vector lies in both the xy- and xz-planes and is perpendicular to the y-axis.
- (0, 2, -1): This vector lies in both the yz- and xz-planes and is perpendicular to the x-axis.
Both of these vectors are perpendicular to both (1, -2, 1) and (1, -2, -1). And as you pointed out, any scalar multiple of these vectors would also work.
You are correct, and I apologize for the mistake. I inadvertently used the wrong vectors when calculating the cross product. Here is the correct solution:
We have two vectors, (3, -6, 3) and (-2, 4, 2), and we want to find two vectors that are perpendicular to both of them.
First, let's find the cross product of (3, -6, 3) and (-2, 4, 2):
(3, -6, 3) x (-2, 4, 2) = ((-6)(2)-(3)(4), (3)(-2)-(3)(2), (3)(4)-(-6)(2)) = (-12, -12, 0)
We can see that the resulting vector has a zero in the z-component, so any vector perpendicular to (-12, -12, 0) must have a z-component of zero.
One possible vector is (6, -6, 0), which lies in the x-y plane and is perpendicular to the z-axis. Another possible vector is (1, 1, 0), which also lies in the x-y plane and is perpendicular to the z-axis.
Therefore, two vectors that are perpendicular to both (3, -6, 3) and (-2, 4, 2) are (6, -6, 0) and (1, 1, 0), or any scalar multiple of these vectors.