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# Find the volume of the parallelepiped with adjacent edges $PQ, PR$, and $PS$.$P (-2, 1, 0)$ , $Q (2, 3, 2)$ , $R (1, 4, -1)$ , $S (3, 6, 1)$

## 16 units $^{3}$

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Welcome back to another cross product problem where we're trying to find the volume of a parallel pipe, ed that have adjacent points, P Q R and S. Where we can calculate three vectors used for our triple product calculation as PQ pr and P s. So then PQ is just q minus p. That's two minus negative too. Three minus one to minus zero. Pr is ar minus P. That'll be 1- negative too. Four minus one and -1 0. And lastly ps will be three minus negative too six minus one. One minus zero. Now, the textbook gives us a nice formula for volume as the absolute value of a dot. With the vector be cross C. But instead of calculating across product and then a dot product, we can just plug in a B and C directly into our matrix here. So we can write for two, two, three, three negative one And 551 and then calculate the cross product. How we normally would With the values for two and 2 replacing I. J. and K. This gives us when we ignore the first column three times one minus negative one times five. And normally we multiply this by I but this time we don't have an eye here, we have four minus and we ignore our second column three times one minus negative one times five. That's one minus negative. One times five. All multiplied by two. Us. Same idea, ignore the third column three times 5 -5 times three. All multiplied by two. Since we don't have any eyes jay's or K's, this isn't a vector, this is a scalar and so we can just add everything up. So we're looking at three minus negative five times four minus three minus negative five times two plus 15 minus 15. Two. Putting this all together, we're looking at 32 minus 16 plus zero, 16 units cubed. Using the triple product method. Thanks for watching

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