plugin-ReviewProbEx1-Prob4

plugin-ReviewProbEx1-Prob4 - Math 255 Winter 2010 Review...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 255 Winter 2010 Review problems for exam 1 Problem 4: Prove that for vectors x = x1 , x2 , x3 and y = y1 , y2 , y3 , |x + y ||x − y | ≤ |x|2 + |y |2 where |x| denotes the magnitude of the vector x. Solution: Using v · v = |v |2 for any vector v , we can write |x + y | = (x + y ) · ( x + y ) = x · x + 2x · y + y · y = |x|2 + |y |2 + 2x · y |x − y | = (x − y ) · ( x − y ) = x · x − 2x · y + y · y = |x|2 + |y |2 − 2x · y then (|x|2 + |y |2 + 2x · y ) (|x|2 + |y |2 − 2x · y ) √ (a − b)(a + b) = a2 − b2 since |x + y ||x − y | = = (|x|2 + |y |2 )2 − 4(x · y )2 ≤ (|x|2 + |y |2 )2 = |x|2 + |y |2 Hence |x + y ||x − y | ≤ |x|2 + |y |2 1 ...
View Full Document

Ask a homework question - tutors are online