# If $\vec a$ and $\vec b$ are such that $|\vec a\times \vec b|=2$, then find $[\vec a~~\vec b~~(\vec a\times \vec b)]$

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?

• Hint: Draw a picture. You have a parallelepiped with base area = 2. What is the height? Apr 15 at 18:14

$$[\vec x ~ ~ \vec y ~~ \vec z] = \vec x\cdot(\vec y \times \vec z) = \vec y\cdot(\vec z \times \vec x) =\boxed{\vec z\cdot(\vec x \times \vec y)}$$

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

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

• (+1) All too easy Apr 15 at 19:13
• Thank you :) @MarkViola
– Ak.
Apr 16 at 1:39
• Ok that works, but I should get the ans by the way I was going too right? Apr 17 at 18:42
• @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}
– Ak.
Apr 18 at 2:18