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
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:
$(2\vec a +\vec b)(2b^2\vec a -(\vec a.\vec b)\vec a +a^2\vec b -2(\vec a. \vec b)\vec b)$
$= 4a^2b^2-2(\vec a.\vec b)a^2 +2a^2(\vec a.\vec b) -4(\vec a.\vec b)^2 +2b^2(\vec a.\vec b)-(\vec a.\vec b)^2+a^2b^2- 2(\vec a.\vec b)b^2$
$=5a^2b^2-5(\vec a.\vec b)^2$