Linear Algebra 3b2: Decomposition with Geometric Vectors 2

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My favorite maths lecturer on YouTube!

jamesspacer

These videos are great. The systematic and step-by-step approach to teaching this is very helpful. I just finished a course in linear algebra and was looking to fill in some gaps that occurred along the way as we had and enormous amount of material to go over but only one semester in which to learn it. These courses are fantastic as a supplement and or as a thorough step-by-step into the world of linear algebra. Thanks again for these videos!

EdKnudsen

A 3rd way to arrive at d is d =a + 3(b-a) since vector b-a appears to be pointing toward d from the tip of vector a. Similarly e = 2*(b-a).

elaproderi

0:57 : example #1
2:04 : example #2 a
4:02 : example #2 b
7:29 : example #3
8:07 : example #4

antonellomascarello

Great lectures ....step by step !For Vector d ...I would just say that it is negative C ( we just found c), plus vector b.Which I figured out from watching these videos.

richt

@Math The Beautiful. I found c=2(a - b) -b = 2a - 3b. Because, "c" is in the direction of "a-b" and 2 times "a-b" will only give us a vector parallel to "c" . So to go upto "c" we have to subtract "b" from the parallel vector to "c" and thus get c=2a -3b . Is it not so, or am i wrong somewhere?

amarparajuli

Would it have been fair to use the information obtained from c as a linear combination of a and b to subtract c from b to find d? As in: since c = 2a - 2b, b - c = b - (2a - 2b) = b - 2a + 2b = -2a + 3b = 3b - 2a? Is this a bad habit in the making or am I actually understanding the point?

TripedalTroductions

Method is too complicated. Instead of trying to express to C as a function of A and B it is far simpler to express A as a function of B and C and then solve algebraically for C. Clearly A = B + 0.5*C so solving for C = 2*(A - B). E then equals - C or 2*(B - A). Then D clearly equal E + B = or 3*B - 2*A. And F = - A.

Jnglfvr

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