Finding a vector perpendicular to two vectors

In summary, the teacher tried to teach us how to do the cross product of vectors V x W, but it was not very helpful because she only showed us an extremely simple example and didn't give us any help with finding the perpendicular vector. We were supposed to just multiply the vectors like scalar numbers, but we were having trouble because the vectors made a down-ward facing V shape which was at a slant. We were able to solve the problem by using a vector that was perpendicular to both V and W.
  • #1
Kevin2341
57
0

Homework Statement


"A vector perpendicular to both vectors V & W
Vector V = 3i + 2j - 2k
Vector W = 4i -3j + k


Homework Equations


Cross product equation



The Attempt at a Solution


The way our teacher tried to teach us was to do the cross product of vectors V x W
Which supposedly equals (area of parallelogram)n-hat (or carrot, what ever you call unit vectors).

But the only example my teacher did this was for an extremely simple set of vectors dealing only with the i and j components and no k component. So basically she went and did V x W for this extremely simple vector problem, and when she chose her perpendicular vector to both V and W, she took a shortcut and just choose k to be the perpendicular vector, which makes perfect sense because it is so simple, but it doesn't help us at all when it comes to trying to do more complex variations of this problem.
 
Physics news on Phys.org
  • #2
Vector V = 3i + 2j - 2k
Vector W = 4i -3j + k

I think you can just multiply them as you would scalar numbers, just keeping in mind
i × j = k
j × k = i
k × i = j
i × i = j × j = k × k = 0

so you would get
(3i + 2j - 2k)(4i -3j + k)
12(0) + (3i)(-3j) + ... + (-2)(0)

I hope I'm remembering this correctly and not providing you with wrong information
 
  • #3
Oh also for the work I have done so far...

I calculated the area of my parallelogram created by the two vectors to be ||V x W||, so its a scalar value, which makes sense. But where I am getting stuck is where to begin with finding the perpendicular vector, I can't just choose a vector going off in a straight component direction because the vectors sort of make a down-ward facing V shape which is at a slant (left side is 2 units higher than right side). So I know that this vector is shooting off in some direction. I can imagine it really easy in my head, its just being able to come up with an actual vector that I'm stuck with, because I have nothing to work with, at least that I know of.
 
  • #4
smk037 said:
Vector V = 3i + 2j - 2k
Vector W = 4i -3j + k

I think you can just multiply them as you would scalar numbers, just keeping in mind
i × j = k
j × k = i
k × i = j
i × i = j × j = k × k = 0

so you would get
(3i + 2j - 2k)(4i -3j + k)
12(0) + (3i)(-3j) + ... + (-2)(0)

I hope I'm remembering this correctly and not providing you with wrong information

Hmm, okay so the vector you came up with is 14i - 6j - 2k, which can be reduced to 7i - 3j - k

But as I remember, is a vector is perpendicular to another vector, their dot product = 0. Using this vector you just calculated, neither V or W = 0 when DP'ed with the perpendicular vector (lets call it P)
 
  • #5
Don't forget, the vector product is not commutative: a x b = - b x a. So

i × j = k, j × i = -k
j × k = i, k × j = -i
k × i = j, i × k = -j

ehild
 
  • #6
ehild said:
Don't forget, the vector product is not commutative: a x b = - b x a. So

i × j = k, j × i = -k
j × k = i, k × j = -i
k × i = j, i × k = -j

ehild

Okay, I got to ask, are you guys doing the Cross Product here? Because I haven't ever seen these things before. For cross products we use determinates

so you have
|..i...j...k...|
|xi(1) yj(1) zk(1)|
|xi(2) yj(2) zk(2)|

= (yj(1)zk(2) - yj(2)zk(1))i - (xi(1)zk(2) - xi(2)zk(1))j - (xi(1)yj(2) - xi(2)zk(1))k
 
  • #7
Yes they are doing cross products. You can do them several ways (your way is correct too).
 

1. How do I find a vector perpendicular to two given vectors?

To find a vector perpendicular to two given vectors, you can use the cross product. Take the two given vectors and calculate their cross product, which will result in a vector that is perpendicular to both of the given vectors.

2. Can there be more than one vector perpendicular to two given vectors?

Yes, there can be an infinite number of vectors that are perpendicular to two given vectors. This is because the cross product of two vectors can result in a vector with different magnitudes and directions.

3. What is the relationship between the dot product and finding a vector perpendicular to two vectors?

The dot product of two vectors can be used to determine if the two vectors are perpendicular to each other. If the dot product is equal to zero, then the two vectors are perpendicular. This can be used as a check when finding a vector perpendicular to two given vectors.

4. Can a vector be perpendicular to itself?

No, a vector cannot be perpendicular to itself. In order for two vectors to be perpendicular, they must have a 90 degree angle between them. A vector cannot have an angle with itself, so it cannot be perpendicular to itself.

5. Is there a specific order in which I need to input the two given vectors when finding a vector perpendicular to them?

Yes, the order of the two given vectors does matter when finding a vector perpendicular to them using the cross product. The first vector should be used as the first operand and the second vector should be used as the second operand in the cross product calculation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
821
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
8K
Back
Top