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Numerade Educator



Problem 11 Easy Difficulty

Find the vector, not with determinants, but by using properties of cross products.

$ (j - k) \times (k - i) $



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Video Transcript

Welcome back to another cross product problem where we try and take the cross product of these vectors without doing any of the math. And the usual method. Just the properties so properties of cross products tells us that if we have a binomial type problem like this, where we have j minus K, crossed with another vector. That's the same thing as J cross. The other vector, K minus I minus K. Cross the other vector. And sure enough, that actually holds going the other way as well, jay cross k minus I. It's the same thing as J Cross K minus, J. Cross I similarly K cross k minus I is the same thing as K. Cross K minus A negative becomes a plus. Okay, Cross I. And so we just need to find the cross products of each of these factors. 1x1. Let's mark down I in black hair, this is I, J in blue, this is jay and K and red, this is K. And so if we want to look at the cross product of J cross K. Then we can use the right hand rule with our right hand fingers pointing in the direction of J. Curling towards K. That leaves our thumb pointing in the direction of their cross product. Which would be I similarly J Cross I If we take a right hand or their fingers pointing at J and curl them towards I. There are some will be pointing in the direction of the cross product which will be straight down or negative. Okay, K Cross K. Well, if we notice if our fingers are already pointing in the direction of Kay, they can't curl towards K. Uh So any vector cross itself or any vector cross, anything parallel to itself Is always going to be zero. And lastly, K cross I if our fingers are pointing at K and curling towards I and our thumb is pointing in the direction of J. And so our final answer is this is helps. This is the victor. I. Okay plus K plus K plus J. Same thing Or the Vector 1, 1, 1. And if we wanted, we could verify that by writing everything out in full and then using our cross product math. That would be a bit of a mess. This is nice and clean. Thanks for watching.