Vector triple product Questions in English

(B) Given $\vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0}$, we have $\vec{b} = -\vec{a} \times (\vec{a} \times \vec{c})$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$, we get $\vec{b} = -[(\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}] = (\vec{a} \cdot \vec{a})\vec{c} - (\vec{a} \cdot \vec{c})\vec{a}$.
Since $|\vec{a}| = 1$, $|\vec{a} \cdot \vec{a}| = 1^2 = 1$. Thus, $\vec{b} = \vec{c} - (\vec{a} \cdot \vec{c})\vec{a}$.
Taking the magnitude squared on both sides: $|\vec{b}|^2 = |\vec{c} - (\vec{a} \cdot \vec{c})\vec{a}|^2 = |\vec{c}|^2 + (\vec{a} \cdot \vec{c})^2 |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c})(\vec{c} \cdot \vec{a}) = |\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2$.
Given $|\vec{b}| = 1$ and $|\vec{c}| = 2$, we have $1^2 = 2^2 - (\vec{a} \cdot \vec{c})^2$, so $(\vec{a} \cdot \vec{c})^2 = 3$.
Since $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos \theta = 1 \cdot 2 \cos \theta = 2 \cos \theta$, we have $(2 \cos \theta)^2 = 3$, which means $4 \cos^2 \theta = 3$.
$\cos^2 \theta = \frac{3}{4} \implies \cos \theta = \frac{\sqrt{3}}{2}$ (for acute angle $\theta$).
Therefore, $\theta = \frac{\pi}{6}$.

(C) Using the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
Given $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b}}{2}$, we have $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b}$.
Rearranging the terms: $(\vec{a} \cdot \vec{c} - \frac{1}{2})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = 0$.
Since $\vec{b}$ and $\vec{c}$ are non-coplanar, they are linearly independent. Therefore, their coefficients must be zero.
Thus, $\vec{a} \cdot \vec{b} = 0$.
Since $\vec{a}$ and $\vec{b}$ are non-zero vectors and their dot product is $0$, the angle between them is $\frac{\pi}{2}$.

(D) Given $\vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0}$.
Using the vector triple product formula $\vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}$.
Since $|\vec{a}| = 1$, we have $\vec{a} \cdot \vec{a} = 1$.
So, $(\vec{a} \cdot \vec{c})\vec{a} - \vec{c} + \vec{b} = \vec{0}$, which implies $\vec{b} = \vec{c} - (\vec{a} \cdot \vec{c})\vec{a}$.
Taking the magnitude squared on both sides: $|\vec{b}|^2 = |\vec{c}|^2 + (\vec{a} \cdot \vec{c})^2 |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c})(\vec{c} \cdot \vec{a})$.
Since $|\vec{b}| = 1$, $|\vec{c}| = 2$, and $|\vec{a}| = 1$, we get $1 = 4 + (\vec{a} \cdot \vec{c})^2 - 2(\vec{a} \cdot \vec{c})^2$.
$1 = 4 - (\vec{a} \cdot \vec{c})^2$, so $(\vec{a} \cdot \vec{c})^2 = 3$.
Thus, $|\vec{a}||\vec{c}| \cos \theta = \pm \sqrt{3}$, where $\theta$ is the angle between $\vec{a}$ and $\vec{c}$.
$1 \cdot 2 \cdot \cos \theta = \sqrt{3} \implies \cos \theta = \frac{\sqrt{3}}{2}$.
Therefore, $\theta = \frac{\pi}{6}$.