Vector perpendicular to two vectors

[aigerimzh]

Homework Statement

A=(i+j-k) B=(2i-j+3k)
Need to find vector perpenducilar to them

Homework Equations

The Attempt at a Solution

I've solved by finding A*B and then divided it to (A*B), the result was W=2i/$\sqrt{38}$-5j/$\sqrt{38}$-3k/$\sqrt{38}$. Is it correct?

 

[HallsofIvy]

What do you mean by A*B? I know scalar product, dot product, and cross product of vectors but none of those use "*". And what is the difference between "A*B" and "(A*B)"? If you mean |A*B|, the length of the vector, I see no reason to divide. The problem did not ask for a unit vector.

 

[aigerimzh]

I mean AxB, and (AxB) means length of AxB. Do mean that my answer is not correct?

 

[HallsofIvy]

No, I meant that I did not understand your answer. Yes, AxB is perpendicular to A and B and so satisfies the condition of the problem. And so AxB/|AxB| is a unit vector perpendicular to both A and B. But the problem did not ask for a unit vector.

 

[rwisz]

Your solution is partially correct. To find a vector perpendicular to two vectors, you can use the cross product. In this case, the cross product of A and B would be a vector that is perpendicular to both A and B.

To find the cross product, you can use the formula: A x B = (A_y*B_z - A_z*B_y)i - (A_x*B_z - A_z*B_x)j + (A_x*B_y - A_y*B_x)k

Plugging in the values for A and B, we get: (1*3 - (-1)*1)i - (1*3 - (-1)*2)j + (1*(-1) - (-1)*2)k = 4i + 1j - 1k

Therefore, the vector (4, 1, -1) is perpendicular to both A and B. You can verify this by taking the dot product of this vector with both A and B, which should result in 0.

So in summary, your solution is correct in terms of finding a vector perpendicular to A and B, but the method you used (dividing A*B by (A*B)) is not the most accurate way. Instead, you can use the cross product to find a vector that is perpendicular to both A and B.