# Vectors: Find the length and direction of u * v

• Nov 14th 2010, 11:43 AM
tintin2006
Vectors: Find the length and direction of u * v
Sorry I meant x not * :)
I need to find the length and direction of u x v, and I have:
u = i x j
v = j x k
So first I will find:

$\displaystyle u \times v = \left[ {\begin{array}{ccc} i & j & k \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} } \right] = i - j + k$

Then the length is 1 and direction is i - j + k.
Am I doing this correctly?
• Nov 14th 2010, 12:24 PM
skeeter
Quote:

Originally Posted by tintin2006
I need to find the length and direction of u * v, and I have:
u = i * j
v = j * k
So first I will find:

$\displaystyle u * v = \left[ {\begin{array}{ccc} i & j & k \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} } \right] = i - j + k$

Then the length is 1 and direction is i - j + k.
Am I doing this correctly?

I assume you mean * to indicate the cross product of two vectors

$\displaystyle u = i \times j = k$

$\displaystyle v = j \times k = i$

so ...

$\displaystyle u \times v = k \times i = j$
• Nov 15th 2010, 03:47 AM
HallsofIvy
Quote:

Originally Posted by tintin2006
I need to find the length and direction of u * v, and I have:
u = i x j
v = j x k
So first I will find:

$\displaystyle u x v = \left[ {\begin{array}{ccc} i & j & k \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} } \right] = i - j + k$

For "$\displaystyle \times$" use "\times" not "x".

Quote:

Then the length is 1 and direction is i - j + k.
Am I doing this correctly?
I find it difficult to understand what you are asking. You used "*" at one point and $\displaystyle \times$ at another. Are they both intended to be the cross product? Also, you say that u= i x j= k and v= j x k= i but then you have i+ j and j+ k instead. Yes, $\displaystyle (i+ j)\times(j+ k)= i- j+ k$ but that has length $\displaystyle \sqrt{1^2+ (-1)^2+ 1^2}= \sqrt{3}$, not 1.
• Nov 15th 2010, 05:41 PM
tintin2006
Sorry and thanks for the tip.
I meant cross product, I didn't notice that * could mean dot product :)
Why does skeeter's solution is so different from mine? Are they the same thing?
Quote:

$\displaystyle u = i \times j = k$

$\displaystyle v = j \times k = i$

so ...
$\displaystyle u \times v = k \times i = j$