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.
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$\begingroup$ what is the meaning of (1^r)? Cross product? $\endgroup$– userAug 1, 2018 at 21:11
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$\begingroup$ it is the cross product of i^r not 1^r $\endgroup$– suyol854Aug 1, 2018 at 21:13
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$\begingroup$ 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. $\endgroup$– amdAug 2, 2018 at 3:02
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$\begingroup$ @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/… $\endgroup$– userSep 6, 2018 at 20:26
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1 Answer
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HINT
We have
$$\vec i \times \vec r=\begin{vmatrix}\vec i&\vec j&\vec k\\1&0&0\\x&y&z\end{vmatrix}=-z\vec j+y \vec k$$
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$\begingroup$ The gradient of that would be equal to (0,-1,1) right? $\endgroup$– suyol854Aug 1, 2018 at 21:30
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$\begingroup$ @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 $\endgroup$– userAug 1, 2018 at 21:34
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$\begingroup$ In this context, $\nabla$ isn’t the ordinary gradient operator of vector calculus. You’re being asked to find the differential of a bivector. $\endgroup$– amdAug 1, 2018 at 22:49