Maximum value of $|(\vec {a} \times\vec {b} ).\vec{c}|$ is $\frac{A}{10}$ Let $\vec {a}$ and $\vec {b}$ be unit vectors. If $\vec{c}$ is a vector such that 
$\vec{c}+(\vec{c} \times \vec{a})=\vec {b}$ then maximum value of $|(\vec {a} \times\vec {b} ).\vec{c}|$ is $\frac{A}{10}$, the find $A$. 
Note: Here |.| represents absolute value
Could someone give me as how to solve this problem. I am not getting how to use given data
 A: $$|(\vec {a} \times(\vec{c}+(\vec{c} \times \vec{a}) ))\cdot\vec{c}|=|(\vec {a} \times\vec{c}+\vec {a} \times(\vec{c} \times \vec{a} ))\cdot\vec{c}|=|(\vec {a} \times\vec{c})\cdot\vec{c}+(\vec {a} \times(\vec{c} \times \vec{a} ))\cdot\vec{c}|$$
The first quantity in the last equality vanishes by orthogonality. Then we can use the BAC-CAB identity to evaluate the triple vector product.
$$=|(\vec {a} \times(\vec{c} \times \vec{a} ))\cdot\vec{c}|=|(\vec {c}(\vec{a}\cdot\vec{a}) -\vec{a}(\vec{a}\cdot\vec{c}))\cdot\vec{c}|=\left||c|^2 -(\vec{a}\cdot\vec{c})^2\right|\le A/10$$
Since $|a|$=1, the maximum value is where $\vec{a}\cdot\vec{c}=0$. Then $|c|^2=A/10$, or $A=10|c|^2$.
The condition $|b|=1$ gives
$$|(\vec{c}+(\vec{c} \times \vec{a})| = 1$$
$$|c|^2+(\vec{c} \times \vec{a})\cdot(\vec{c} \times \vec{a}) = 1$$
Using another identity, $(A\times B)\cdot(C\times D)=(A\cdot C)(B\cdot D)-(B\cdot C)(A\cdot D)$,$^1$ we get
$$|c|^2+|c|^2|a|^2-(\vec{a}\cdot\vec{c})^2 = 1$$
$$|c|^2+|c|^2-(\vec{a}\cdot\vec{c})^2 = 1$$
Again, at the maximum $\vec{a}\cdot\vec{c}=0$, so $|c|^2 = 1/2$.
Thus $A=5$.

$^1$ I just recognized that this identity is essentially the "epsilon killer" of index notation.
A: Here is another solution. Choosing an appropriate orthogonal coordinate frame, we may assume that $\vec{a}$ and $\vec{b}$ are of the form $\vec{a} = (1, 0, 0)$ and $\vec{b} = (\cos\theta, \sin\theta, 0)$. Letting $\vec{c} = (x, y, z)$, the given condition satisfies
$$ (x, y, z) + (0, -z, y) = (\cos\theta, \sin\theta, 0). $$
Solving this relation gives $x = \cos\theta$ and $y = -z = \frac{1}{2}\sin\theta$. Now since
$$ |(\vec{a} \times \vec{b})\cdot\vec{c}| = |(0, 0, \sin\theta)\cdot \vec{c}| = \tfrac{1}{2}\sin^2\theta, $$
it follows that the maximum possible value of $|(\vec{a} \times \vec{b})\cdot\vec{c}|$ is $\frac{1}{2}$. Equating this with $\frac{1}{10}A$ gives $A = 5$.
