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How do I find the dot product in case of problem 1a exercise 9.3 (advance engineering mathematics Greenberg) Note: this is for self study.

Diagram in url https://i.stack.imgur.com/rfWsj.jpg

Got it   $\sin(60^\circ) = \|u\| / \|v\| \quad \Longrightarrow \quad \|v\| = 10 / \sqrt{3}$

$$u\cdot v = \|u\| \|v\| \cos(150^\circ) = 5 \times \frac{10}{\sqrt{3}} \times \frac{- \sqrt{3}} 2 = -25$$

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  • $\begingroup$ $$u\cdot v = ||u||||v||\cos\alpha,$$ where $\alpha$ is the angle between... $\endgroup$
    – 5xum
    Sep 13, 2015 at 8:36
  • $\begingroup$ Please clarify since the |v| is not given in this case $\endgroup$ Sep 13, 2015 at 8:42
  • $\begingroup$ $||v||$ is not given, but you have a right angle triangle which you can use. $\endgroup$
    – 5xum
    Sep 13, 2015 at 8:45
  • $\begingroup$ Looks about right, yes. $\endgroup$
    – 5xum
    Sep 13, 2015 at 8:51
  • $\begingroup$ 5xum thanks for pointing me in right direction. $\endgroup$ Sep 13, 2015 at 8:57

1 Answer 1

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You have $|v|=5\sec 30$, so $u\cdot v=5\times5\times \frac{2}{\sqrt{3}}\times\cos 150=-25$

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