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Welcome back to another cross products problem. This time
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we're trying to determine if the four points given our
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co planner. Using a triple product method, the
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way we can do that is first find three vectors
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in this plane starting at the same origin. So
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we could look at for example, A B,
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A C and a D. And then using the
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triple product method, calculate the volume of the corresponding
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parallel pipe pad. If that volume is zero that
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means they're parallel pipe at a squished flat and therefore
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the points are all co planner. If not,
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if the parallel pipe it has volume, then they're
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not co cleaner. So A B is just b
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minus a three minus one,-1-3, six
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minus two. A C is C minus A.
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That's 5-1, two minus three. In 0
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-2 and 80 Is 3-1. Six minus three
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negative four minus two in the triple product method.
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Instead of doing this formula as a cross product and
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then a separate thought product, we can just plug
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in A. B and C directly into this matrix
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here. So there's a there's be and there.
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See. And then we can calculate the cross product
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same as normal except instead of I. J and
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K will have these three numbers here. Usually what
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we would do is ignore the first column and look
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at negative one times negative six minus-2-3.
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one times negative six minus-2 times three. And
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instead of I We're going to multiply by two,
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Dennis. Then we ignore our second column four times
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negative six minus negative two times two, four times
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negative six-20 Times two. And instead of being
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multiplied by J, We're multiplying by-4 plus.
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And then we ignore our 3rd column four times three
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. It is one times 2, all multiplied by
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four. Since we don't have any I Js or
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k's. This is not a vector, but just
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a scalar. So let's add everything up. Six
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minus negative six. Thank you. 12 times two
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minus negative, 24 minus negative four. I'll be
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-20 times minus four and then lastly 12 minus negative
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two times four. Plugging this all in. We're
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getting 24 minus 80 plus 56 Which is indeed zero
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. Therefore, all of these points are actually co
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planner. Thanks for what?