💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!



Numerade Educator



Problem 38 Hard Difficulty

Use the scalar triple product to determine whether the points $ A (1, 3, 2), B (3, -1, 6), C (5, 2, 0) $, and $ D (3, 6, -4) $ lie in the same plane.


The given points are coplanar

More Answers


You must be signed in to discuss.

Video Transcript

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