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enter image description hereenter image description hereenter image description hereIf $\vec{a}$ and $\vec{b} $ are vectors in space given by $\vec{a} = \frac{\hat{i}-2\hat{j}}{\sqrt{5}}$ and $\vec{b} = \frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}$ , then value of $ (2\vec{a}+\vec{b}).[(\vec{a}\times\vec{b})\times(\vec{a}-2\vec{b})]$ is

(A) ${2}$

(B) ${3}$

(C) ${5}$

(D) ${13}$

Okay, so this a question from my book which I have been attempting for the last 2 hours. I solved it in the traditional way by solving each part separately and then combining them together, but still, the answer came out to be wrong. this is an MCQ question so I am guessing that there must be a way of solving this question in much less time and with more accuracy. can somebody please help me out with this question. Thanks Regards

calculations

additional_calculations

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closed as off-topic by Saad, Michael Rybkin, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, blub Apr 17 at 16:17

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. $\endgroup$ – 5xum Apr 16 at 9:10
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    $\begingroup$ Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. Even if the question is closed, you can still edit it, and we will vote to reopen it. $\endgroup$ – 5xum Apr 16 at 9:10
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    $\begingroup$ Note: This is the second PSQ of this user so far with a title almost identical to the first one. $\endgroup$ – Saad Apr 16 at 9:15
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    $\begingroup$ yeah sure, I will edit and upload the calculations i did $\endgroup$ – gucci Apr 16 at 9:16
  • $\begingroup$ I didn't know what to name the title, can you please suggest one so that I can edit it $\endgroup$ – gucci Apr 16 at 9:17
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HINT:

Do your calculation with $a$ and $b$ first and after simplifying what you have as much as possible substitute the values of $a$ and $b$

You'll get that $(2\vec{a}+\vec{b}).[(\vec{a}\times\vec{b})\times(\vec{a}-2\vec{b})]=5a^2b^2 - 5(\vec a.\vec b)^2$

Also, $a^2=||\vec a||^2=1$, $b^2=1$ and $(\vec a. \vec b)=0$

So the right answer will be $(c)$ $5$

Rules you'll use:

$(1)$ $\vec u × (\vec a +\vec b)=\vec u × \vec a + \vec u × \vec b$

$(2)$ $(\vec u × \vec v)×\vec w=(\vec u. \vec w)\vec v - (\vec v. \vec w)\vec u$

$(3)$ $\vec a. \vec b=\vec b. \vec a$

$(4)$ $\hat i. \hat i= \hat j. \hat j=1$

$(5)$ $\hat i. \hat j=0$

$(6)$ $\vec a.\vec b=a _xb_x +a_yb_y + a_zb_z$

$(7)$ $a^2=\vec a. \vec a= ||\vec a||.||\vec a||.\cos (0)= ||\vec a||^2$

What you'll get is:

$(2\vec a +\vec b)(2b^2\vec a -(\vec a.\vec b)\vec a +a^2\vec b -2(\vec a. \vec b)\vec b)$

$= 4a^2b^2-2(\vec a.\vec b)a^2 +2a^2(\vec a.\vec b) -4(\vec a.\vec b)^2 +2b^2(\vec a.\vec b)-(\vec a.\vec b)^2+a^2b^2- 2(\vec a.\vec b)b^2$

$=5a^2b^2-5(\vec a.\vec b)^2$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Asaf Karagila Apr 16 at 12:50

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