See the image below Projection has two parts: i The direction where you're projecting onto. Tags: Vectors Projection. Related What is meant when we say that a differential takes on a certain value? Conceptual question on eigenvectors in quantum mechanics Showing that 4D rank-2 anti-symmetric tensor always contains a polar and axial vector Why is acceleration directed inward when an object rotates in a circle?
Properties of the dot product Help with vector cross product identity Alternative expression of acceleration in vector form Problems regarding components of velocity Covariant vs contravariant vectors. So we can view it as the shadow of x on our line l. That's one way to think of it. Another way to think of it, and you can think of it however you like, is how much of x goes in the l direction? So the technique would be the same.
You would draw a perpendicular from x to l, and you say, OK then how much of l would have to go in that direction to get to my perpendicular? Either of those are how I think of the idea of a projection. I think the shadow is part of the motivation for why it's even called a projection, right? When you project something, you're beaming light and seeing where the light hits on a wall, and you're doing that here. You're beaming light and you're seeing where that light hits on a line in this case.
But you can't do anything with this definition. This is just kind of an intuitive sense of what a projection is. So we need to figure out some way to calculate this, or a more mathematically precise definition. And one thing we can do is, when I created this projection-- let me actually draw another projection of another line or another vector just so you get the idea. If I had some other vector over here that looked like that, the projection of this onto the line would look something like this.
You would just draw a perpendicular and its projection would be like that. But I don't want to talk about just this case.
I want to give you the sense that it's the shadow of any vector onto this line. So how can we think about it with our original example? In every case, no matter how I perceive it, I dropped a perpendicular down here. And so if we construct a vector right here, we could say, hey, that vector is always going to be perpendicular to the line.
And we can do that. I wouldn't have been talking about it if we couldn't. So let me define this vector, which I've not even defined it. What is this vector going to be? If this vector-- let me not use all these.
We know we want to somehow get to this blue vector. Let me keep it in blue. That blue vector is the projection of x onto l. That's what we want to get to. Now, one thing we can look at is this pink vector right there. What is that pink vector? That pink vector that I just drew, that's the vector x minus the projection, minus this blue vector over here, minus the projection of x onto l, right?
If you add the projection to the pink vector, you get x. So if you add this blue projection of x to x minus the projection of x, you're, of course, you going to get x.
We also know that this pink vector is orthogonal to the line itself, which means it's orthogonal to every vector on the line, which also means that its dot product is going to be zero. So let me define the projection this way.
The projection, this is going to be my slightly more mathematical definition. The projection onto l of some vector x is going to be some vector that's in l, right? I drew it right here, this blue vector. I'll trace it with white right here. Some vector in l where, and this might be a little bit unintuitive, where x minus the projection vector onto l of x is orthogonal to my line.
So I'm saying the projection-- this is my definition. I'm defining the projection of x onto l with some vector in l where x minus that projection is orthogonal to l. Asked 6 years, 11 months ago. Active 2 years, 8 months ago. Viewed 16k times. John Trentnor John Trentnor 83 1 1 gold badge 1 1 silver badge 3 3 bronze badges. Evidently, you don't use that phrase to mean the same thing as "the vector projection", so what do you mean by that term?
I don't see what there is to explain. Show 3 more comments. Active Oldest Votes. See the image below Projection has two parts: i The direction where you're projecting onto. Add a comment. Ross Millikan Ross Millikan k 26 26 gold badges silver badges bronze badges. I have no clue why this would be of interest. OP has just provided the definition his class is using, and it's a scalar, not a vector. It should make sense that the component of any vector pointing in the same parallel direction is simply equal to the magnitude of the vector.
Similarly, there is no component of a vector that points in the perpendicular direction, hence the scalar projection in this direction is equal to 0. To solve this question, we should first recognize that the algebraic projection of one vector in some given direction is another way of describing a scalar projection. As required by the question, our answer has been simplified and expressed in decimal form to the nearest hundredth two decimal places.
Scalar projection can also be used in examples related to geometric systems. The cube shown has sides lengths of 4 4 1 7. For this question, we are asked to find the scalar projection of one vector onto another. These two vectors are the diagonals of a cube.
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