Problem 46 Hard Difficulty

(a) Let $ P $ be a point not on the plane that passes through the points $ Q, R $, and $ S $. Show that the distance $ d $ from $ P $ to the plane is
$$ d = \frac{\mid a \cdot ( b \times c) \mid}{\mid a \times b \mid} $$
where $ a = \vec{QR} , b = \vec{QS} $, and $ c = \vec{QP} $.
(b) Use the formula in part (a) to find the distance from the point $ P (2, 1, 4) $ to the plane through the points $ Q (1, 0, 0) , R (0, 2, 0) $, and $ S (0, 0, 3) $.

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WZ

Wen Z.

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Video Transcript

Welcome back to another cross product problem. This time we're looking at what happens if we want to find the distance from a point. Let's call it P to a plane. We know we can define a plane by three different points. Let's call these Q, R and S. And so if we define a couple of vectors, let's say a between Q and r. Be between Q and S. And say see between Q and r. Point. Then what we can do is try and figure out the distance from the point to the plane. Using some things that we know about. Triple products. Specifically the volume of the triple product of the parallel pipe ed between a B and C is defined as the magnitude of across B times the magnitude of C. Cosign theta. And so rearranging this a little bit. If we write C dot a cross B divided by the magnitude of a cross B, then this is going to be the magnitude of C. It's the length of that vector C times Cosine Theta where theta is the angle between our vector and perpendicular. This is theater right here. Now, since cosign Theta is adjacent over hypotenuse, that means the length of this perpendicular lines. The distance from the point of the plane, going to be the magnitude to see times. Cosine Theta. We can use this if we want to calculate the distance from say point P 214 to the plane divided by Q, R and S. In order to do that first, we need to calculate projectors defining the plane. So A is Q. R. AR -Q Sierra -1 To -00 0 and Q. S. That's s minus cues. Again zero minus one, zero minus zero, three minus zero. We'll also need to see that's the vector QP p minus q is two minus one. One minus zero, four minus zero. So if we want to calculate the magnitude of c dot a cross B, we can do that in one step using our triple product. If we plug in C. A and B into our matrix here, I'll point out this isn't the only way we can do this. Um But we'll talk about another way in just a second. So let's plug in 114 -1-0 And -103. And using the formula from our textbook, remember we ignore the first column And look at two times 3 minus zero times zero, Going to be 6 0. And normally we would multiply by I We're actually gonna multiply by one minus and ignore the second column -1 times three zero times negative one. Eight of three minus zero times not J but times one again us. And then we ignore the third column negative one times zero minus two times negative one will be zero minus negative too, Which is 0-plus two times four. And since we don't have any eyes jay's or k's this is not a factor but just a number six times one minus negative three, That's plus three times 1 Plus two times 4. And so we're looking at six plus three plus eight is 17. The other thing we need from our formula remember is the magnitude of a cross B. And so we can get that by calculating the cross product of A and B. Once again we're looking at six minus zero, I minus negative three minus zero jay plus zero minus negative too. Okay, Giving us the vector six three two. If you want to calculate the magnitude of a Crosby, that'll just be the magnitude of 63 two. Which is the square root of six squared Plus three squared plus two squared or The square root of 36 plus nine plus four. And that's the square root of 49 or just seven. Since we determined that the distance is the ratio of the two numbers that we just found, we get the distance is let's go back to that cross product Or the triple product with 17 over seven. I said earlier, this wasn't the only way we could have found this. Since we already know the cross product a Crosby, you could have the magnitude of that and then found the magnitude of C. Got a cross B. Just using the vector C. And a dot product. That would avoid having to do the cross product twice. Thanks for watching.