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

1. ## 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?

2. 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 ...

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

3. 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".

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.

4. 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?
$\displaystyle u = i \times j = k$

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

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

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