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If $\vec B$ is added to $\vec A$, the result is 4.8i + 1.2j. If $\vec B$ is subtracted from $\vec A$ , the result is −3.20 + 5.8j. What is the magnitude of $\vec A$?

Am I supposed to subtract the two, so 8i - 4.6j? But then how would I find A's magnitude? To find magnitude, I would need to multiply direction, length, and a random choice of either sin or cos. BUt I can't do that with the data I have.

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  • $\begingroup$ If a random choice is allowed, then the magnitude of $A$ must be $4.5564367$. $\endgroup$ – Yves Daoust Sep 21 '16 at 13:56
  • $\begingroup$ Isn't random choice the formula? I had a previous problem where I just chose to use sin randomly and it marked it right. What's the formula? $\endgroup$ – user366783 Sep 21 '16 at 13:58
  • $\begingroup$ I don't see how a random choice of either $\sin$ or $\cos$ would be part of the solution to a question that asked, "What is the magnitude of the vector?" I suppose there must be something about the other question that you have not told us. Or maybe you just made a lucky guess. $\endgroup$ – David K Sep 22 '16 at 12:50
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$$(A + B) + (A - B) = 2A$$ $$||2A|| = 2||A||$$

Edit: I missed the part about the magnitude. $$||A|| = \sqrt{A\cdot A}$$

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Hint: Let $A=xi+yj$ and $B=ai+bj$. Then $$(x+a)i+(y+b)j=4.8i+1.2j$$ $$(x-a)i+(y-b)j=-3.2i+5.8j$$ Equating the compoenents (i.e. $x+a=4.8$, $y+b=1.2$ etc.) will give you four equations to solve for four variables.

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  • $\begingroup$ How would I equate the components? And what would I do with the four variables? $\endgroup$ – user366783 Sep 21 '16 at 13:44
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    $\begingroup$ See, $x+a=4.8$ and $x-a=-3.2$. Adding both the equations give you $2x=0.4$ which implies that $x=0.2$. Similarily $y+b=1.2$ and $y-b=5.8$. Adding these two equations gives you $2y=7.0$ which implies that $y=3.5$. Thus $A=0.2i+3.5j$ and so the magnitude of $A$ is $\sqrt{0.2^2+3.5^2}$. $\endgroup$ – model_checker Sep 21 '16 at 14:20

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