If $|a|=2$ and $|b|=3$ and $a\cdot b=0$, then determine $$( a\times( a\times( a\times( a\times b))))$$ where $a$ and $b$ are vectors.
What I am doing wrong here?
We know
$$|a\times b|^2 + |a\cdot b|^2 = |a|^2|b|^2$$ where $\times$ is the cross product and $\cdot$ is the dot product.
Then after simplification $$(a\times b) = 6$$
Then I can put $(a\times b) = 6$ at the end of the question and the question will be $$a\times( a\times( a\times( 6))))$$
Isn't the answer not equal to $0$ as $a\times a =0$.
Where I am wrong?
You should use the triple product in this exercise
$$u\times (v\times w)=(u\cdot w)\,v-(u\cdot v)\,w$$
Applying for $u=v=a$ and $w=b$ you get
$a\times (a\times b)=\underbrace{(a\cdot b)}_0\,a-\underbrace{(a\cdot a)}_{|a|^2}\,b=-4\,b$
And you can iterate the processus since you now have $a\times (a\times (-4b))=(-4)^2\,b=16\,b$