What is the cross product of $(1,0,-2)$ and $(0,1,-2)$?

Question

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?

Answer 1

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

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

Answer 2

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

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 $\mathbf i \times \mathbf k = -\mathbf j$.)

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

Answer 3

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 $\left\{i,j,k\right\}$.