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The required values is of $$\vec a. (\vec b\times (\vec a\times \vec b))$$ $$=\vec a.((\vec b.\vec b).\vec a-(\vec b.\vec a).(\vec b))$$ $$=\vec a.(b^2\vec a-(\vec b. \vec a).\vec b)$$ $$=2a^2b^2+a^2b^2$$ $$=2a^2b^2$$

Also $ab\sin x=2$

How should I proceed?

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    $\begingroup$ Hint: Draw a picture. You have a parallelepiped with base area = 2. What is the height? $\endgroup$ Apr 15 at 18:14
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$[\vec x ~ ~ \vec y ~~ \vec z] = \vec x\cdot(\vec y \times \vec z) = \vec y\cdot(\vec z \times \vec x) =\boxed{\vec z\cdot(\vec x \times \vec y)}$

Taking $\vec x = \vec a$, $\vec y = \vec b$, $\vec z = \vec a \times \vec b$ you get,

$[\vec a ~ ~ \vec b ~~ \vec a \times \vec b] = (\vec a \times \vec b) \cdot(\vec a \times \vec b) = |\vec a \times \vec b|^2 = 4$

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    $\begingroup$ (+1) All too easy $\endgroup$
    – Mark Viola
    Apr 15 at 19:13
  • $\begingroup$ Thank you :) @MarkViola $\endgroup$
    – Ak.
    Apr 16 at 1:39
  • $\begingroup$ Ok that works, but I should get the ans by the way I was going too right? $\endgroup$
    – Aditya
    Apr 17 at 18:42
  • $\begingroup$ @Aditya Your method also works, but there's a mistake in your last step. $\begin{align}\\\vec a\cdot(b^2\vec a - (\vec b\cdot\vec a)\vec b) &= b^2a^2 -\color{blue}{(\vec a\cdot\vec b)^2} \\&= a^2b^2-a^2b^2\cos^2\theta \\&= a^2b^2\sin^2\theta \\&= |\vec a\times\vec b|^2 = 4\end{align}$ $\endgroup$
    – Ak.
    Apr 18 at 2:18

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