How to solve cross and scalar product equation knowing length and angle

Question

We have two vectors $a$ and vector $b$. they are both non zero vectors and the angle between them is $\frac{\pi}{6}$. Their lenghts are connected with equation $|b| = \sqrt{3}|a|$

We need to solve this equation (*...scalar product, $\times$... cross product):

I find it very hard since I know that scalar product is not associative.

My try: $$ (x*a+x*b)*a + x \times b = 2a\times b + 3|a|^{2}b\\ (x*a)*a+(x*b)*a + x \times b = 2a\times b + 3|a|^{2}b$$ If I solve the distance of $2a\times b$, I get nowhere. I do not know hot to continue.

The solution is $x = 2a+ \frac{2}{9}b+ 2a\times b $

Answer 1

The given data gives $\vec a. \vec b=a^2$. Let $\vec x= u\vec a+v \vec b+ w (\vec a \times \vec b).$ where $\vec a$ and $\vec b$ are non-collinear. Inserting it in the equation $$[\vec x.(\vec a+\vec b)]\vec a+\vec x \times \vec b= 2\vec a\times \vec b + 3a^2\vec b.$$ $$ [\{u \vec a + v \vec b+ w(\vec a\times \vec b)\}.(\vec a + \vec b)].\vec a+ [u\vec a+v \vec b+ w (\vec a \times \vec b)]\times \vec b=2 \vec a \times \vec b +3a^2 \vec b$$ In this Eq. One has to compare the co-efficients of the bases $\vec a$, $\vec b$ and $\vec a \times \vec b$ to get the scalars $u,v,w$. One can make use of $(\vec a \times \vec b)\times \vec b= (\vec a. \vec b)\vec b-b^2 \vec a.$ and $\vec a.(\vec a \times \vec b)=0.$

I hope to come back.

Answer 2

Use the fact that $a$, $b$, and $2a\times b$ is a basis. Let $$x=\alpha a+\beta b+\gamma (2a\times b).$$ Multiply out and compare coefficients; use that $3\|a\|^2=\|b\|^2$ and $2\langle a,b\rangle=\|b\|^2$.