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I'm a little confused on this problem. The Question asks to find the cross product, $a \times b$.

$$a = \langle 6,0,-2\rangle\text{ and }b = \langle 0,8,0\rangle$$

The formula that is given in my text book is $a \times b = \langle a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1\rangle$

And I get $\langle 16, -8, 48\rangle$

However, the book says that the answer is $16i + 48k$? How are they getting that? Could someone tell me what I doing wrong, and also what exactly the cross product is supposed to be. I understand that is is a vector product, but can someone explain that in more details?

Thanks in advance.

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"And I get <16, -8, 48>" - but you didn't say how you got it. Without saying how you did things, it's hard to point where you went wrong. – J. M. Sep 8 '12 at 3:40
The result for the middle term (i.e. the $j$ term) is -(6*0 - (-2)*0) which is 0. How you result in -8 is beyond me as no two factors of 8 are in the original vectors. – Tpofofn Sep 8 '12 at 3:40
up vote 1 down vote accepted

If you look under Matrix Product in Wikipedia, the cross product is $$\left |\begin {array} {c c c} i&j&k\\6&0&-2\\0&8&0 \end {array}\right |$$ The zeros eliminate a lot of terms, leaving $-i(-2\cdot 8)+k(6\cdot 8)$.

Added: to take a $3 \times 3$ matrix determinant you multiply the elements in each of the diagonals down and to the right (including the broken ones) and subtract the products of the diagonals up and to the right. So this is $i\cdot 0 \cdot 0 + j \cdot (-2) \cdot 0 + k \cdot 6 \cdot 8 -i \cdot (-2)\cdot 8 - j\cdot 6 \cdot 0 - k \cdot 0 \cdot 0$ The particular terms are a bit harder to follow because of the several zeros, but it should be clear.

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Yea, I got it, and this is the correct answer. I just wish you had explained how to do this problem in more detail. I arrived at the correct answer, after some time. – Avalon-96 Sep 8 '12 at 19:59

You haven't given the formula from your textbook, but the computation for the middle, or $j$, term, should be $(-2*0-6*0)j=0j$, giving $<16,0,48>$ as your text says. How'd you get that $-8,$ when the only products of two entries of your vectors are $16$ and $48$, up to sign?

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With $a = \left(6,0,-2\right)$ and $b = \left(0,8,0\right)$, $$ \begin{eqnarray} \left(a_1,a_2,a_3\right) \times \left(b_1,b_2,b_3\right) &=& \left(a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1\right) \\ &=& \left(16,0,48\right). \end{eqnarray} $$ Wikipedia has a decent explanation of the cross product.

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By definition, the product of vectors $a= (a_x, a_y, a_z)$ and $b = (b_x, b_y, b_z)$ equals $$a \times b = \det \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a_x & a_y & a_z\\ b_x & b_y & b_z \end{pmatrix},$$ where $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ is the standard basis.

So we have $$a \times b = \det \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 6 & 0 & -2\\ 0 & 8 & 0 \end{pmatrix} = 16\, \mathbf{i} + 0\, \mathbf{j} + 48\, \mathbf{k}.$$

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