j x k , vectors

**dazygrl02** · May 27th 2010, 10:37 AM

Okay, I turned in a piece of homework the other day and my teacher said that I got one wrong. I was asked to solve for j x k or <0, 1, 0> x <0, 0, 1>. I said that the answer was i (<1, 0, 0>). But, try as I might that is the only answer that I am getting. Am I wrong? Is she wrong? Thanks.

**Plato** · May 27th 2010, 10:57 AM

Originally Posted by dazygrl02

Okay, I turned in a piece of homework the other day and my teacher said that I got one wrong. I was asked to solve for j x k or <0, 1, 0> x <0, 0, 1>. I said that the answer was i (<1, 0, 0>). But, try as I might that is the only answer that I am getting. Am I wrong? Is she wrong? Thanks.

Here is a complete listing:
$\begin{array}{lll}<br /> {i \times j = k} & {j \times k = i} & {k \times i = j} \\<br /> {j \times i = - k} & {k \times j = - i} & {i \times k = - j} \\<br /> <br /> \end{array}$

**Soroban** · May 27th 2010, 07:14 PM

Hello, dazygrl02!

You are right . . . she is wrong.

$\text{Given: }\,\begin{array}{ccc}j &=& \langle 0,1,0\rangle \\ k &=& \langle 0,0,1\rangle \end{array} \quad \text{ Find: }j \times k$

I said that the answer was: . $i \:=\:\langle1, 0, 0\rangle$

$j \times k \;=\;\left|\begin{array}{ccc} i & j & k \\ 0&1&0 \\ 0&0&1\end{array}\right| \;=\;i(1-0) - j(0-0) + k(0-0) \;=\;i$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

There is a diagram for multiplying these unit vectors.

. . $\begin{array}{ccccc} & & i \\ & \nearrow & & \searrow \\ k & & \leftarrow & & j \end{array}$

If the multiplication is clockwise, the product is positive.

. . $\begin{array}{ccc} i \times j &=& k \\ j \times k &=& i \\ k \times i &=& j \end{array}$

If the multiplication is counterclockwise, the product is negative.

. . $\begin{array}{ccc}j \times i &=& -k \\ k \times j &=& -i \\ i \times k &=& -j \end{array}$

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