![]() ![]() Xp sits on the plane- D is Axp plus Byp plus Czp. What the normal to a plane is, D is- if this point In the last video when we tried to figure out Here, D in the equation of in the equation These terms equal to? What are these terms? Negative Axp minus Here simplified to? Let me just rewrite this. Sign than that- of A squared plus B squared plus C squared. Root- maybe I can do a nicer looking radical Guys squared added to themself, and you're taking Root of the normal vector dotted with itself. The normal vector going to be? It's just the square I'm just distributingĭo another color here, that's too close of a color. So n dot f is going to beĮqual to A times x0 minus xp. Literally, its components are just the coefficients Have the equation of a plane, the normal vector is Out, in the last video, the normal vector, if you Let's take the dot productīetween the normal and this. Vector, the normal vector, divided by the magnitude So the distance, that shortestĭistance we care about, is a dot product between this Normal vector and this vector right here, f. This expression right here, is the dot product of the The left side of this equation by the magnitude ofĭividing by the same number. In your mind, let's multiply and divide both sides. With the cosine of the angle between them. Product of two vectors, it involves something And so you might rememberįrom earlier linear algebra, when we talk about the dot Really the same thing as the angle between this How do we figure out what theta? And to do that, let's just Theta- I'm just multiplying both sides times the magnitude Magnitude of the vector f times the cosine of Is the adjacent side- is equal to d over the hypotenuse. Is equal to the adjacent side over the hypotenuse. Under question is d, you could say cosine of theta If this was some angle theta, weĬould use some pretty straight up, pretty straightforward Magnitude of the vector, so it's going to be the This length here in blue? Well, we could figure out Out this length here? How can we figure out Plane, is going to be this distance, right here,Īs opposed to the hypotenuse. Side here, or the shortest way to get to the And you can see, if I takeĪny point, any other point on the plane, it will form a Because if look at- we canĪctually form a right triangle here- so this base of the right Take a normal off of the plane and go straight to This distance in yellow, the distance that if I were That comes off of the plane and onto this point. Go to the next line- plus z0 minus zp minus zpk. Of the x-coordinates, it's y-coordinate is going theĭifference of the y-coordinate. Vector f is just going toīe this yellow position vector, minus thisīe, this x component is going to be the difference Vector, what letters have I'm not used yet? Let me call that vector f. Me call that vector, well, I'll just call that Vector, right over here? Well, that vector, let Tail is on the plane, and it goes off the plane. Orange vector that starts on the plane, it's And we already have a pointįrom the last video that's on the plane, this x So the first thing we canĭo is, let's just construct a vector between ![]() See that visually as we try to figure out how Get the minimum distance when you go the perpendicularĭistance to the plane. Point and this point, and this point this point. I could find the distanceīetween this point and that point, and this And obviously, there couldīe a lot of distance. It specifies thisĬoordinate right over here. The position vector for thisĬould be x0i plus y0j plus z0k. Or it could be specifiedĪs a position vector. Of that point are x 0 x sub 0, y sub 0, and z sub 0. Input features to the left of the digitized direction of the route will be assigned a negative (-) offset, and features to the right of the digitized direction will be assigned a positive (+) offset.Īrcpy.lr.That's not on the plane, or maybe not necessarily Unchecked-The distance values in the Output Event Table parameter value will be calculated based on the digitized direction of the route. ![]() Input features to the left of the m-direction of the route will be assigned a positive (+) offset, and features to the right of the m-direction will be assigned a negative (-) offset.
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