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Quantum Mechanics in Simple Matrix Form: chapter 5 of (2-28)

Vectors
Vectors

I use “x” for vectors that are pointing farthest to the left (that is \pi radians) and I use “y” for the other vector, or rather, a vector not pointing farthest to the left :)

5-1.

Find x_1 y_1 + x_2 y_2 + x_3 y_3

(a)

Given:

x_1 = 0 \newline x_2 =4 \cdot cos( \frac{\pi}{4} ) \newline x_3 = 4 \cdot sin( \frac{\pi}{4})

y_1 = 0 \newline y_2 = 5 \newline y_3 = 0

> x_1 y_1 + x_2 y_2 + x_3 y_3=

0 \cdot 0+ 4 \cdot \cos( \frac{ \pi}{4}) \cdot 5 + 4 \cdot \sin(\frac{\pi}{4}) \cdot 0 =

4 \cdot \cos( \frac{ \pi}{4}) \cdot 5 = 14.1421356

_

(b)

Given:

x_1 = 0 \newline x_2 = 0 \newline x_3 = 4

y_1 = 0 \newline y_2 = 5 \newline y_3 = 0

>x_1 y_1 + x_2 y_2 + x_3 y_3 =

0 \cdot 0 + 0 \cdot 5 + 4 \cdot 0 =

0
_

(c)

Given:

x_1 = 0 \newline   x_2 = 4 \cos(\frac{3\pi}{4}) \newline     x_3 = 4 \sin(\frac{3\pi}{4}) 

y_1 = 0 \newline   y_2 = 5 \newline   y_3 = 0

>x_1 y_1 + x_2 y_2 + x_3 y_3 =

0 \cdot 0 + 4\cos(\frac{3\pi}{4})\cdot 5 + 4\sin(\frac{3\pi}{4}) \cdot 0 =

4\cos(\frac{3\pi}{4})\cdot 5 = -2.82842712 
_


(d)

Given:

x_1 = 0 \newline   x_2 = -4 \newline     x_3 = 4 \sin(\frac{3\pi}{4}) 

y_1 = 0 \newline   y_2 = 5 \newline   y_3 = 0

>x_1 y_1 + x_2 y_2 + x_3 y_3 =
0 \cdot 0 + -4 \cdot 5 + 0 \cdot 0 =
-4 \cdot 5 = -20
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This entry was posted on February 27, 2011 at 9:31 pm and is filed under Quantum Mechanics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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