All, I came across this question... I could only think of solving it using a graphical method (or rather, tip to tail method as in physics) at the moment... Does anyone have any idea if it could be solved by non-graphical method?
Show that for all non-zero vectors, a and b,
|a + b| ≤ |a| + |b|
Thanks
Hi,
Good question and not an easy one!
You may wish to refer to the details on this site:
http://ltcconline.net/greenl/courses/203/Vectors/vectors.htm
In essence, we need to do the following:
1. show that |a . b| <= |a| |b| (Cauchy-Schwartz inequality);
2. square LHS expression and then use the result in step 1 to prove RHS expression.
Thanks!
Cheers,
Wen Shih
Thanks!
And.... would a graphical solution suffice in an exam?
Posting the solution from the website
The proof is this is quite tricky. We let x be a scalar and note from property 1 that
0 < (xu + v) . (xu + v)
= u . u x2 + 2 u . v x + v . v
Since the dot product produces a scalar, the above equation is a quadratic in x. A quadratic that is always positive has nonpositive discriminant. Hence
(2 u . v)2 - 4(u . u)(v . v) < 0
dividing by 4 gives
(u . v)2 - (u . u)(v . v) < 0
or
(u . v)2 < (u . u)(v . v)
Taking the square root of both sides produces the result.
Hi,
I believe such a question will not appear in an exam, unless there are guided steps for students to follow. Btw, the Cauchy-Schwartz inequality is very specialised knowledge that our syllabus does not include, but HK maths syllabus cover & assess it quite comprehensively. Thanks!
Cheers,
Wen Shih
Actually it's from AJC tutorial, modified from RJC 2004 prelim question...
This is the first part of the question... It's leads to proving magnitude of the sum of vectors being smaller than the magnitude of each individual vector by induction
The simplest way to solve it, so far as I could see, is by graphical method. Was wondering if there's any non-graphical method that is in H2 syllabus... Thanks!
Hi,
It's probably set as a challenge for able students. Thanks!
Cheers,
Wen Shih
Hi,
AJC maths tutorials may be quite ridiculous at times, haha!
As educators, we should help students to focus on key concepts first. If the student is real competent, we can then direct his/her energy to up his/her intellect :) Thanks!
Cheers,
Wen Shih
Actually on second thought.... we can use the definition of dot product to solve this
Because a.b = |a||b| cos θ
And cos θ ≤ 1
Hence a.b ≤ |a| |b|
(a+b).(a+b) = a.a + 2 a.b + b.b
= |a|² + 2 a.b + |b|²
So
a.b ≤ |a| |b|
2 a.b ≤ 2|a| |b|
|a|² + 2 a.b + |b|² ≤ |a|² + 2|a| |b| + |b|²
(a+b).(a+b) ≤ ( |a| + |b| )²
|a + b|² ≤ ( |a| + |b| )²
Square root both sides gives us
|a + b| ≤ |a| + |b|
The brain is indeed clearer in the morning :D
Note this inequailty apply to any two complex numbers too
|a + b| ≤ |a| + |b|
|a + b|^2 = (a+b)(a+b)* = (a+b)(a*+b*) = a.a* + b.b* + a.b* + b.a*
= |a|^2 + |b|^2 + a.b* + (a.b*)*
= |a|^2 + |b|^2 + 2Re (a.b*)
≤ |a|^2 + |b|^2 + 2|a.b*| = |a|^2 + |b|^2 + 2|a||b| = (|a| + |b|)^2
jiaxing
Hi,
Good work, eagle and jiaxing2!
Now, we have the triangle inequality for real numbers, vectors and complex numbers! Can submit paper to conference le, haha!
Cheers,
Wen Shih