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Problem 30 Hard Difficulty

(a) Find a nonzero vector orthogonal to the plane through the points $ P, Q, $ and $ R $, and (b) find the area of triangle $ PQR $.

$ P (0, 0, -3) $ , $ Q (4, 2, 0) $ , $ R (3, 3, 1) $


A.$<-1,-7,6>\quad$ or $\quad<1,7,-6>$
B.$A r e a=\frac{\sqrt{86}}{2}$

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Video Transcript

Welcome back to another cross products problem Were given three points in three d. Space and we're asked to find victor Orthogonal to the plane created by those three points. So one of the things we can do is given these three points. We can find two vectors, say PQ and Pr And since we have two vectors in the plane, the cross product between them is going to be perpendicular to that plane. So PQ is just q minus P. That's going to be 4 0. It's for to minus zero, which is two and zero minus negative three, Which is just three. Similarly Pr is ar minus P three minus zero, three minus zero, one minus negative three. We can plug these vectors into our matrix here for two, three and three three, four. And then use the technique outlined in our textbook defined PQ Cross pr So remember we ignore the first column of this matrix And then look at two times 4 And it's three times 3, two times 4 -3 times three I minus. Then we ignore the second column four times four And it's three times 3 it's for -3 times three J. Plus. And then we ignore the third column And we'll look at four times 3 -2 times three three -2 times three. Okay, simplifying this a little bit two times four is eight, three times 3 is nine. So we're looking at negative one in our first coordinate four times 4 is 16, three times 3 is nine, 15 -97. So we're looking at negative seven And then four times 3 is 12, 2 times three is 6, -6. His six of course the negative of this vector would also be perpendicular to the plane. And so either of these answers or any scalar multiples of them will work. Now, what if we wanted to find the area of this triangle peak? You are? Well, it turns out you can do that just by looking at the absolute value of the cross product divided by two. Let's write that out. So the area of P. Q. The area of triangle. There we go. He Q. R. Is just going to be the magnitude of PQ cross pr that leave us the area of a parallelogram. And then we divide that by two to get the area of that triangle. And so this is just going to be the square root of the first component squared plus the second component squared plus the third component squared, All divided by two. And so we're looking at the square root of one plus 49 plus 36 All divided by two, giving us the square root of 86 over to as the area of triangle P Q R. Thanks for watching.

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