Answered: 1 (|x+y[? – |x]? – \y|*) 2

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Use properties of the dot product to simplify the expression.

Draw a picture illustrating the resulting identity.

Expert Solution

Step 1

step:-1

Since, we know for vector $\vec{v} = (v_{1}, v_{2}) a n d \vec{u} = (u_{1}, u_{2})$

$|\vec{v}| = \sqrt{v_{1}^{2} + v_{2}^{2}} \Rightarrow {|\vec{v}|}^{2} = (v_{1}^{2} + v_{2}^{2}) = (s c a l a r) - - - - (1) N o w, \vec{v} . \vec{v} = (v_{1}, v_{2}) . (v_{1}, v_{2}) = (v_{1}^{2} + v_{2}^{2}) N o t e : - (v_{1} i + v_{2} j) . (v_{1} i + v_{2} j) = (v_{1}^{2} + v_{2}^{2}) s o, b y (1), {|\vec{v}|}^{2} = \vec{v} . \vec{v} S i m i l a r l y, f o r {|v + u|}^{2}$

Step:-2

Given that $\frac{1}{2} ({|x + y|}^{2} - {|x|}^{2} - {|y|}^{2})$

Solving this expression using above results we have

$\frac{1}{2} ({|x + y|}^{2} - {|x|}^{2} - {|y|}^{2}) = \frac{1}{2} ((x + y) (x + y) - (x . x) - (y . y)) = \frac{1}{2} ((x . x) + x . y + y . x + (y . y) - (x . x) - (y . y)) = \frac{1}{2} (x . y + y . x) Since, dot product is commutative as result of product is a scalar So, x . y = y . x t h e n = \frac{1}{2} (x . y + x . y) = x . y \Rightarrow \frac{1}{2} ({|x + y|}^{2} - {|x|}^{2} - {|y|}^{2}) = x . y$

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