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I know this is an easy question. I've just done it and got:

$(2,-2,1)$

But I was reading my lecturer's notes on Surface Integrals and there was a point where this came up within a question. He wrote:

$(2,2,1)$

And used that through the entirety of the question.

Is my answer wrong?

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    Your answer is wrong, the dot product of what you got with the second vector is not zero. Remember, if you take the cofactor expansion (i,j,k), you need to negate the second term. – 2017-02-21

3 Answers 3

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We can calculate the cross product by computing the determinant of the following matrix

$$ \mathbf{u}\times \mathbf{v}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ u_1 & u_2 & u_3\\ v_1 & v_2 & v_3 \end{array}\right|$$

Therefore we have

$$ \begin{align} (1,0,-2)\times (0,1,-2)&=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 0 & -2\\ 0 & 1 & -2 \end{array}\right|\\ &=\left|\begin{array}{cc} 0 & -2\\ 1 & -2 \end{array}\right|\mathbf{i} - \left|\begin{array}{cc} 1 & -2\\ 0 & -2 \end{array}\right|\mathbf{j} + \left|\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right|\mathbf{k}\\ &=(0+2)\mathbf{i}-(-2-0)\mathbf{j}+(1-0)\mathbf{k}\\ &= 2\mathbf{i}+2\mathbf{j}+\mathbf{k}\\ &= (2,2,1) \end{align}$$

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Using the cyclic rule for evaluating the cross product of the unit basis vectors

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it is very simple to show that you've got the wrong sign on your second component. (Note: in case you're not aware of the cyclic rule, it says that taking a cross product in the direction of the arrows yields a positive. And going against the arrows gives a negative. So for instance $\mathbf j \times \mathbf k = \mathbf i$ while $\mathbf i \times \mathbf k = -\mathbf j$.)

$$\begin{align}(\mathbf i -2\mathbf k)\times (\mathbf j-2\mathbf k) &= \mathbf i\times \mathbf j -2\mathbf i\times \mathbf k -2\mathbf k\times \mathbf j +4\mathbf k\times \mathbf k \\ &= \mathbf k+2\mathbf j+2\mathbf i + \mathbf 0 \\ &= 2\mathbf i+2\mathbf j+\mathbf k\end{align}$$

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    This is two of your recent answers that have fancy stuff going on behind the scenes (I'm thinking about the "derivative as a best linear approximation" answer). Nice! – 2017-02-21
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    Glad you like it. :-) – 2017-02-21
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You can calculate it with

$$ det\left( \begin{matrix} i & j & k \\ 1 & 0 & -2 \\ 0 & 1 & -2 \\ \end{matrix}\right) $$

where you apply Laplace expansion theorem to the firts row.

In this case the result will be $2\cdot i-(-2)\cdot j+1\cdot k$, written as $(2,2,1)$ with respect to the basis $\left\{i,j,k\right\}$.