# Calculate $\nabla (\mathbf i\wedge\mathbf r)$

I was working out a tutorial and the question was like this. "Consider r = xi + yj + zk, work out $\nabla (\mathbf i \wedge \mathbf r)$." I am not sure what value should I take for $\mathbf i$. According to the answer it should be equal to -i.

• what is the meaning of (1^r)? Cross product?
– user
Aug 1, 2018 at 21:11
• it is the cross product of i^r not 1^r Aug 1, 2018 at 21:13
• Actually, you’ll have to tell us what $\mathbf i$ and the other symbols mean. Based on the supposed answer, $\nabla$ is not the usual vector calculus gradient operator since the gradient of a vector is a rank-two tensor. I suspect that $\wedge$ is not really the usual cross product, either. On the other hand, if $\mathbf i$ and friends are meant to be the basis vectors and $\wedge$ and $\nabla$ have their usual geometric algebra/calculus meanings, then the answer is missing a factor of $2$. Surely there are definitions earlier in the tutorial for all of these symbols—share them with us.
– amd
Aug 2, 2018 at 3:02
• @user7075815 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– user
Sep 6, 2018 at 20:26

HINT

We have

$$\vec i \times \vec r=\begin{vmatrix}\vec i&\vec j&\vec k\\1&0&0\\x&y&z\end{vmatrix}=-z\vec j+y \vec k$$

• The gradient of that would be equal to (0,-1,1) right? Aug 1, 2018 at 21:30
• @user7075815 usually the gradient is on a scalar field, here we are calculating the gradient of a vector which is a matrix, refer to math.stackexchange.com/q/1565075/505767
– user
Aug 1, 2018 at 21:34
• In this context, $\nabla$ isn’t the ordinary gradient operator of vector calculus. You’re being asked to find the differential of a bivector.
– amd
Aug 1, 2018 at 22:49