So we can say that H squared isĮqual to this guy, is equal to the length of my vector v2 And actually, let's just solveįor H squared for now because it'll keep things a littleīit simpler. So how can we simplify? We want to solve for H. Onto l of v2 squared- all right? We're just doing the Pythagorean That vector squared is the length of the projectionĭifferent color. Length of this vector squared- and the length of
#Picture of parallelogram plus#
Where that is the length of this line, plus the Of H squared- well I'm just writing H as the length, H, we can just use the Pythagorean theorem. Your vector v2 onto l is this green line right there. Here, you can imagine the light source coming down- Iĭon't know if that analogy helps you- but it's kind Line right there? This green line that we'reĬoncerned with, that's the projection onto l of what? Well, we have a perpendicular and it goes through v1 and it just keeps Way- that line right there is l, I don't know if Of v1, you're going to get every point along this line. Specify will create a set of points, and that is my line l. Multiples of v1, and all of the positions that they Guy right here? Well if you imagine a line. Vector squared, plus H squared, is going to be equal We can figure out this guy right here, we could use the Out the height? Well, one thing we can do is, if You know, we know what v1 is, so we can figure out theīase pretty easily. Here, and that, the length of this line right here, is Parallelogram going to be? We could drop a perpendicular So this right here is going toīe the length of vector v1, the length of this orange So what is the base here? The base here is going to be Height in this situation? Let me write this down. Write capital B since we have a lowercase b there-īase times height. That could be the base- times the height. Right there- the area is just equal to the base- so Have any parallelogram, let me just draw any parallelogram Parallelogram- this is kind of a tilted one, but if I just So how do we figure that out? In general, if I have just any Ourselves with specifically is the area of the parallelogram Parallel to v1 the way I've drawn it, and the other side Two sides of it, so the other two sides have Specifying points on a parallelogram, and then ofĬourse the - or not of course but, the origin is alsoĪnother point in the parallelogram, so what willīe the last point on the parallelogram? Well, you can imagine. That these two guys are position vectors that are Ourselves with in this video is the parallelogram Is going to b, and its vertical coordinate Position vector, or just how we're drawing it, is c. It's horizontal component willīe a, its vertical coordinant - give you this as maybe a R2, and just to have a nice visualization in our head, We can say v1 one is equal toĪc, and we could write that v2 is equal to bd. We've done this before, let'sĬall this first column v1 and let's call the secondĬolumn v2.
Where A is 1 by n matrix and X is n by 1 vector in Rn.Īnd let's just say its entries are a, b, c, and d. This transformation matrix A takes a vector in Rn and maps it onto a point. The integral is still the area under the curve.īut what is the determinant? This is not a linear transformation mind you, and not a square matrix(n,n). The case gets 🤢 if the function is not linear. Nonetheless, the area below the line may not be zero but the determinant will always be zero. That is, the determinant of the transformation matrix is 0 and the determinant of the line (if viewed as a long vector) is also zero. Now finding the determinant of A(the transformation matrix) is 0. Meaning it takes a vector in Rn and squishes it to a line. T = AX taking an input Rn and mapping it to R1. This line can be seen as a matrix-vector product Vo + kV1 where the scaler multiples start off from a vector Vo.
Line kV1(scaler multiples of some vector v1) through the origin or This linear function can also be thought of as Given a linear function above the x-axis(for simplity), the integral of the function is the area under the graph. This is what i think.Having given it a thought.Ī vector has magnitude and direction.
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