2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 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 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 at 18:17
5
$\begingroup$

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 $|\vec a-\vec b|$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks!, I dont know how I missed that, smh. $\endgroup$ – riyad malakh May 22 at 16:20
  • 2
    $\begingroup$ It's quite alright. We all make mistakes and learn from them :) $\endgroup$ – sai-kartik May 22 at 16:21
  • 2
    $\begingroup$ @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 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 at 16:44
  • 2
    $\begingroup$ @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 at 16:46
5
$\begingroup$

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 $\mathbb{R}^n$?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – riyad malakh May 22 at 16:20
  • 1
    $\begingroup$ @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 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.