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A problem I'm facing involves finding out the modulus of vectors. I now know how to solve and get the answer for 20. (a), but I am unsure how to do (b). I was thinking it would become |r-2r| = |-r| = |-5|. Is there any rule that would make it -|5| which would result in the answer being 5?

Also, I wanted to know how to solve for part (d) of the question. I know how to solve for (c) but did not understand how the vectors can be drawn for part (d).

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2 Answers 2

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For a)$$|r+5r|=|6r|=6|r|$$

b)$$|r-2r|=|-r|=|r|$$ c)$$|r+s|^2=|r|^2+|s|^2=…$$ d)$$s\cdot(r+s)=s\cdot r+|s|^2=0$$

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for b, there is indeed a rule that is $ |\lambda \times v| = |\lambda|\times |v|$ with $\lambda = -1, |\lambda| = 1$ .

For d, to draw it, you just need to set $v = r + s$. Then you can draw $s$ and $v$, and you know $r = v - s$

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