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Problem 12 Medium Difficulty

Find the distance from $ (4, -2, 6) $ to each of the following.

(a) The $ xy $-plane (b) The $ yz $-plane
(c) The $ xz $-plane (d) The $ x $-axis
(e) The $ y $-axis (f) The $ z $-axis

Answer

(a) 6
(b) 4
(c) 2
(d) 2$\sqrt{10}$
(e) 2$\sqrt{13}$
(f) 2$\sqrt{5}$

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Video Transcript

So for part a of this problem, we're trying to find the distance from the point P with coordinates four minus 26 to the X Y plane. So I've drawn a picture for part A. You have a x y and Z axes, and we have the X Y plane. And I've drawn a point p assholes, Inspector V and this is really we're gonna be trying to find here to find the distance. Well, he wanted to find some vector. That's the shortest vector from the plane to the point and you want to win its magnitude. That's what's gonna give us the distance. So, um, we know that the direction of the specter should be in the same direction as the normal vector to the plane, which I've already written us N since this is the X Y plane. And she just pointed a Z direction. So it's a vector with components 001 But we need to scale this so that it's the right distance so that it ends it p. And so to do this, we're gonna use a vector projection using this vector P had with the components of P. So this vector p hat is is it's a specter of drawn that starts at the origin and ends of the point P. And it has the same components as the point p itself. So the distance should just be the magnitude of this vector v from the plane to pee. That's in this perpendicular direction. And so again, this this will just really be the projection of the P vector onto the normal vector. Go back and see the formula for this. Since N is a unit vector, this will just end up being the matter of the magnitude the absolute value of the dot product of N p. So in this case, we have zeros and the 1st 2 coordinates. So we're just gonna grab this last coordinates six, and that will be our distance. That shouldn't be too surprising, since when x y plane make. I don't care about the X and y coordinates. Um, but the Z coordinate is gonna tell us how high it's It's about the plane. So into the same thing for part B, where we're trying to find the distance from the point to the y Z plane organise a different normal vector in this case Just looking at the y Z plane. The normal vector should point in the ex direction so we can choose this. And then again, we have our point vector, if you like. The vector that points to our point and the distance as we derived last time is just gonna be the absolute value of the stock products. And in this case, we're only gonna be grabbing this first coordinate for a distance off four. And we're into the same thing for part C. Now we're trying to find the distance to the X Z plane so our normal vector will point in the why direction and then, as usual, will form our point. Vector in the distance is just gonna be the absolute value of their dot Products in this case were only grabbing the middle coordinate and taking its absolute value, which is to So let's move on and do these last three parts, these ones are gonna be a little different are actually very different. As instead of finding the distance from a point to a plane, we're finding the distance from a point to a line in particular the three axes. So it's quite a few different ways to do this. I'm gonna be showing how to do this in my favorite way, which uses some coke, one optimization techniques. So we're gonna do first is I've already written a premature ization for the ex access. So the X axis means that the wine Z components are zero shaken. Pramuk tries this line with just some parameter and the first component since we just need to specify some X value. So, um, if we've specified some point on the line, we can find the distance from that point t 00 to the point p using the distance function. So if we choose some tea, our parameter in the distance from that point on the line to the point p will look like this. And we can simplify this after foiling or four minus t squared. And I believe you should get 56 minus 80 plus t squared. So we've done is we've found some formula. Or if you pick a point on the X axis with X coordinate T and you plug it into this little square root over here into D of T, you'll find the distance from that point to pee And so it really trying to dio when we're finding the distance from P to the X axis to this line is we want to minimize this function dysfunction d of t. So if you remember from calculus one, the way you minimize the function of one real variable is by taking the first derivative and setting it equal to zero. So we'll take the first derivative using the chain role. We'll get 1/2 on the bottom. We have our function, and on the top we have the dirt of the inside which will be in minus eight, plus two tea. And then again, when you're trying to find where this has a minimum, we set the sequel to zero, so the bottom isn't gonna come into play. It's also get to note that this is the distance function itself, so it should be positive. Since P doesn't lie on this line, we're not dividing by zero, which is good. And we also ignore the constant out front sensor sitting in equal to zero. You could multiply by two to cancel it out. So it's gonna set this top here equal to zero. We're gonna set minus eight plus two t equals zero. And if we add it to the other side and divide by two, we should find that the distance function is minimized when t equals four. And so what that tells us is that the distance frumpy to the X axis is really just this distance function evaluated AT T equals four. And so you can just plug for in over here to our square root. After simplifying it. I believe you should get to Route 10. Okay, It's nice little application of some Cal Kwan ideas here close to the same thing for party. We're finding the distance from the point to the y axis. So we privatize the Y axis is I've done here, and we're just gonna write the distance from that prescribed point in terms of tea on the Y axis to our point p, which should just look like this. So to save some time, I'll let you do the derivative calculation in solving for zero yourself to check. I believe that the derivatives should be zero when t is equal to minus two. It's okay. So then, just like last time, our distance is just gonna be this function evaluated at this point, which once you do, so I think should give you two. Route 13. Okay, let's do our last one finding the distance to the Z axis. So again we premature eyes are Z axis with just a tea and Z component. We formed the distance function from an arbitrary point on the Z axis. Premature ized by t to the point p which should be given by this expression again. Um, I'll let you work it out and take the derivative sitting in equal to zero. I believe you should get that this happens when t equal six. So the 60.6 is the closest point on the sea. Access to this point p again. Not very surprising. It has the component six s. So then we can find the distance that we're looking for by plugging in six to our distance function. And if you do this, I believe you should get to Route five to revive