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I done some question that was supposed to be in for today, and i've managed to lose the working out for my vector question...

Vectors $u, v$ are known to be orthogonal vectors. Find the missing entry. $$\vec u = \begin{pmatrix} 6 \\ -3\end{pmatrix},\quad \vec v = \begin{pmatrix} -4\\ ?\end{pmatrix}$$

This was the question and i'm almost certain that the answer was -14, but i can't remember and can't think how to do it anymore :@@@@@

Source: http://vvcap.net/db/aQW7u-WrN71ZhXwMoE09.htp

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Perhaps you could share with us what you can remember of your previous solution? I find it very hard to believe that you can't remember anything at all ... –  Old John Nov 11 '13 at 13:23
    
Call the missing coordinate $x$ and solve the equation $u\cdot v=0$. The answer is not $-14$. –  Per Manne Nov 11 '13 at 13:23
    
it had to be equal to 0... –  harry Nov 11 '13 at 13:25
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1 Answer 1

Let $\vec{x}=(x_1,x_2), \vec{y} = (y_1,y_2) \in \mathbb{R}^2$.

$$ \vec{x} \; \; \text{is orthogonal to} \; \; \vec{y} \iff \vec{x} \dot{} \vec{y} = x_1y_1 + x_2y_2 = 0$$

Using this fact, your problem is solved.

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@amWhy why did you change my arrows? –  Chasky Nov 11 '13 at 13:38
    
You had underlined the vectors; I simply changed that to over-line arrows. If you really wanted to underline the vectors, my apologies; you can simply "roll back" to your original answer. –  amWhy Nov 11 '13 at 13:40
    
This is a simple yet precise hint. +! –  DonAntonio Nov 11 '13 at 13:40
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We are so cheap, @amWhy ... –  DonAntonio Nov 11 '13 at 13:44
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thank god for that! i done it last minute! yesss :D and btw i honestly had done it i was up till 3am doing it one night and lost it lol :D cheers –  harry Nov 11 '13 at 19:01
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