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Numerade Educator



Problem 52 Hard Difficulty

Prove that
$ (a \times b) \cdot (c \times d) = \left \|
a \cdot c & \mbox{$ b \cdot c $}\\
a \cdot d & \mbox{$ b \cdot d $}
\end{array} \right \| $


$(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=\left| \begin{array}{cc}{\mathbf{a} \cdot \mathbf{c}} & {\mathbf{b} \cdot \mathbf{c}} \\ {\mathbf{a} \cdot \mathbf{d}} & {\mathbf{b} \cdot \mathbf{d}}\end{array}\right|$

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Video Transcript

Welcome back to another cross product problem, or where we're trying to find the dot product of across B and c. Cross D. And figuring out what this is equal to ideally as a determinant. And the way that we can approach this is by taking see crusty and by writing it as a new vector. Yeah, so we're really looking at a cross B dot e and our cross product identities tells us that this is the same thing as a dot. Be cross E. If we expand this out, this is just a dot, be cross. And I remember E was just see cross D at this point, we can use the helpful identity that we've used in the past that says that a triple cross product is really The 1st, 3rd time, 2nd minus first at second time. Third, let's put that into use. This is a diet. And then, like I said, first times first dot third, that will be he got the time, see minus 1st and 2nd, that will be e dot c. Time's D. Now you'll notice that be dot de is just a number as his be dot C. Therefore, this is really a dot, a number of times see -1. Our Number of Times D. What we're really looking at is a dot c, times this number be dot de minus the number. Be dot c times a dot be. You will notice that this looks kind of like a determinant, meaning we could write this as the determinant of the matrix A dot c. Pierotti. No. And you don't see a dot de. So that we're looking at a dot c b dot de minus b dot c a dot de. And that is what this dot product of cross products is equal to control.