If $a, b$ and $c$ are three vectors with magnitudes $1, 1$ and $2$ respectively and $a \times (a \times c) + b = 0$, then the angle between $a$ and $c$ is

$(b \times c) \times (c \times a) = \dots$

Let $\vec{a}=-5 \hat{i}+\hat{j}-3 \hat{k}$,$\vec{b}=\hat{i}+2 \hat{j}-4 \hat{k}$ and $\vec{c}=(((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to

If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors, then $\frac{\vec{a} \times(\vec{b} \times \vec{a})}{|\vec{a}|^2}$ represents:

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j}$,$\vec{c} = \hat{i}$ and $(\vec{a} \times \vec{b}) \times \vec{c} = \lambda \vec{a} + \mu \vec{b}$,then $\lambda + \mu$ is equal to :-

Statement $(A)$ : If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,then $\vec{a} \times (\vec{b} \times \vec{c}) = 0$.
Reason $(R)$ : If $\vec{b}$ is perpendicular to $\vec{c}$,then $\vec{b} \times \vec{c} = 0$.