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Numerade Educator



Problem 4 Easy Difficulty

What are the projections of the point $ (2, 3, 5) $ on the $ xy $-, $ yz $-, and $ xz $- planes? Draw a rectangular box with the origin and $ (2, 3, 5) $ as opposite vertices and with its faces parallel to the
coordinate planes. Label all vertices of the box. Find the length of the diagonal of the box.




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Video Transcript

All right. So we're gonna begin between plotting the 0.235 do on the X Factor's three on the y and then five on Dizzy What? 23 and then go up. Five. Let's say it's about right there, right. And then we're gonna look at the projections of this point on three planes the X Y plane, B Y Z plane and the XZ plane. So the protections it always makes the access that it's missing, let's say the X Y plane, for example, the projection of the X Y plane just makes this point um z 0.0 since its X and Y we make these zero. So what that means is, this point is 235 we make the zero 230 so we go toe to three zero and it should be directly straight under from the original point. And we do this for the other to access these y Z. That means that the equity zero, But for this one, it is 35 and x zero and then for the XY plane, it is why that is zero. So for XY is two and five. So the point is 205 do 05 all right, and now we connect them well, that's just mean he's on. Then this was straight down. Okay, so there's a point. Um, it's going to raise the little doc, all right? And then we're gonna have to make corners of this rectangular prism. And so it's pretty simple. You just go to the intersection with the Y Axis, for example, so everything 050 that's this 500.30 That's that point, um, connects to that one with X axis. It would be 200 so excellent or y in 00 And then we're coming with that point. And then for the Z, it's 005 at this point, and then that goes over here. So this top faith are it's basically rectangular prison. And then the last part is finding the distance, um, of the diagonal so that the diagonal is from the origin. 2.235 That would be let's say that. Okay, So to find the distance of this line, it used the distance formula for two points. Two points are 235000 So we'll just be taking the square root of the sum of the squares of the court. So let's begin with the X axis. The two X coordinates are two and zero. Okay. And then we'll add that to the difference between the white coordinates now, which is three and zero. And then we'll add that to to Z. All right. Perfect. So then we just go to the next step, which is squares of these four, nine and 25 and you add it and you should get 38. All right, So yeah, things who just summarize really quickly you put the point first and the projections are basically making the third access that isn't part of that plane. Zero So x y that means zero R 00 and then find the distance of the diagonal is just from the origin to the point. Use this handy dandy dis informative and that be it