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I can't seem to understand this question at all. It does not make sense to me.

The question is

Given $\left|\vec a\right| = 3, \left|\vec b\right| = 5$ and $\left|\vec a+\vec b\right| = 7$. Determine $\left|\vec a-\vec b\right|$.

I have tried finding $\left|\vec a+\vec b\right|$ using cosine rule such that $\left|\vec a+\vec b\right| = 7 = 3^2 + 5^2 - 2\cosθ$

Which failed as I clearly am unable to picture this question correctly in my head. If someone could explain this question (or maybe help me sketch it) that's be very helpful, thanks in advance.

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  • $\begingroup$ a+b and a-b are diagonals of a parallelogram with two sides of length 3 and two of length 5. $\endgroup$
    – sykh
    May 22 '20 at 15:56
  • $\begingroup$ Also $\vec a-\vec b=\vec a+(-\vec b)$ . This means that you're turning $\vec b$ by 180 degrees and then adding it to $\vec a$ $\endgroup$
    – sai-kartik
    May 22 '20 at 15:58
  • $\begingroup$ You should learn the parallelogram law if you haven’t already. It comes up in a variety of contexts. $\endgroup$
    – amd
    May 22 '20 at 18:17
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HINT

If you are in, for example, $\mathbb{R}^2$, write $a = (a_x,a_y), b = (b_x,b_y)$ then $$ \begin{split} |a+b|^2 &= a_x^2 + b_x^2 + 2a_x b_x + a_y^2 + b_y^2 + 2a_yb_y\\ |a-b|^2 &= a_x^2 + b_x^2 - 2a_x b_x + a_y^2 + b_y^2 - 2a_yb_y \\ |a+b|^2 + |a-b|^2 &= 2a_x^2 + 2b_x^2 + 2a_y^2 + 2b_y^2 \\ &= 2|a|^2 + 2|b|^2 \end{split} $$ Can you finish and generalize for $\mathbb{R}^n$?

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  • $\begingroup$ I was able to finish the answer! But I'm not sure about my generalization. I found that in R3 it is also 2|a|^2 + 2|b|^2 Since in this case a = (ax,ay,az) and b = (ax,ay,az) Is that correct? $\endgroup$ May 22 '20 at 16:20
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    $\begingroup$ @riyadmalakh yes, that is the general relationship in $\mathbb{R}^n$. Turns out, holds in more complicated vector spaces as well. $\endgroup$
    – gt6989b
    May 22 '20 at 16:41
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You are thinking in the right direction, but have used the wrong formula. The right ones are: $$|\vec a\pm\vec b|^2=|\vec a|^2+|\vec b|^2\pm2|\vec a||\vec b|\cos\theta$$ Since you have $|\vec a+\vec b|$, you can find $\cos\theta$ and then plug in the value to find $|\vec a-\vec b|$

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  • $\begingroup$ Thanks!, I dont know how I missed that, smh. $\endgroup$ May 22 '20 at 16:20
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    $\begingroup$ It's quite alright. We all make mistakes and learn from them :) $\endgroup$
    – sai-kartik
    May 22 '20 at 16:21
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    $\begingroup$ @sai-kartik +1 and it can be argued that this type of an experience makes one absorb the ideas much deeper. $\endgroup$
    – gt6989b
    May 22 '20 at 16:42
  • $\begingroup$ True @gt6989b I've learned this from real-life experience..(part of it from this website) $\endgroup$
    – sai-kartik
    May 22 '20 at 16:44
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    $\begingroup$ @sai-kartik it's been a pleasure seeing your answers throughout the site recently, happy to virtually meet you. Hope you stay around and contribute to the site. All the best for you and yours. $\endgroup$
    – gt6989b
    May 22 '20 at 16:46

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