If a, b, c are non coplanar unit vectors.. Such that $ a \times (b \times c) =\frac{ b + c}{\sqrt{2}}$ prove the angle between a, b is 3/4pi. I really don't want open the thing with vector so i cross both sidrs first with b and then with a I'll get $ c \times ( b \times (a \times (b \times c))) = a \times b \times a \times b \times c = \frac{a \times b \times c}{\sqrt{2}}$ Since cross product is unique $ a \times b \times a = \frac{a}{\sqrt{2}}$ upon dotting both sides with a and rearranging i get the wrong answer ie a = 0. Where am i going wrong? Continuing to error - $- a \cdot (a \times a \times b) = a \times b \cdot (a \times a) $ (a. bx c = c.axb with c = axb)

You must use the rule for triple vector product: $$ a\times(b\times c)=b(a\cdot c)-c(a\cdot b)=b\cos\gamma - c\cos\beta, $$ where $\gamma$ and $\beta$ are the angles formed by $a$ with $c$ and $b$.

Comparing that with $a\times(b\times c)=(b+c)/\sqrt2$ we get then: $$ \cos\gamma={1\over\sqrt2},\quad \cos\beta=-{1\over\sqrt2}. $$