(1) As a max-min problem in one real variable, determine the value of t that min
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Question
(1) As a max-min problem in one real variable, determine the value of t that minimizes the function f(t) = |A tB| where t ( < t < ) It is easier to work with the square f(t) 2 . Be sure to justify this.
(2) Let L be the line defined by the position function F(t) = tB where t ( < t < ). Let M be a point of L which is nearest to A. Use the result of part (1) to verify that the line joining the endpoint of A to M is perpendicular to L.
(3) Prove that the inequality |(A·B)| |A||B| holds for all vectors A, B. Note. In this proof, you shall not use the usual formula for the dot product in terms of the cosine of an angle. Indeed, it is this proof which justifies the formula known to you. Hint: What is the relationship between the discriminant of a quadratic polynomial and the location of its roots?
(4) Prove that |A + B| |A| + |B|. How is this inequality interpreted geometrically?
Question 1 I think I have the correct point. t = (a1b1+a2b2+a3b3)/(b1^2+b2^2+b3^2) (the numbers right after the letters are subscripts). And I have shown how it is a minimum. Its the next questions that stumped me.
Okay so for question 2, I have worked out this problem to where I have dotted Vector MA with F(t) (L) to try to prove that it equals zero. This is where I am having problems. Can anyone help out here?
For questions 3 and 4 I am kinda lost? Not really sure how to prove or explain why the inequality makes sense.
Explanation / Answer
minimize f(t) = |A tB|
I am gong to use <A,B> to represent the inner (dot) product between A and B
|A tB|^2 = <A tB, A tB>
= |A|^2 + t^2|B|^2 - 2t <A,B>
t = <A,B> / |B|^2
2)
show that <A-tB, B> = 0
<A - (<A,B> /|B|^2) B, B>
<A,B> - <A, B> <B,B> / |B|^2
<B,B> = |B|^2
<A,B> - <A, B> = 0
3) prove the Cauchy-Schwarz equation
<A,B> |A| |B|
|A| > 0 for all non 0 A.
|A - tB|^2 = |A|^2 + t^2|B|^2 - 2t<A,B> 0
|A + tB|^2 = |A|^2 + t^2|B|^2 + 2t<A,B> 0
|A|^2 + t^2|B|^2 2t<A,B>
|A|^2 + t^2|B|^2 -2t<A,B>
A|^2 + t^2|B|^2 | 2t<A,B>|
complete the square on the left.
|A|^2+ t^2|B|^2 - 2t|A||B| + 2t|A||B| |2t <A,B>|
(|A| - t |B|)^2 + 2t|A||B|| |2t <A,B>|
there exists a t>0 such that |A| = t|B|
2t|A||B| 2t |<A,B>|
|A||B| |<A,B>|
fyi, a consequence of this (together with fact that the minimum distance in the universe is the plank length) is the Heisenberg uncertainty principle!
4)
|A + B| |A| + |B| this is the triangle inequality
the shortest distance between two points is a straight line.
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