vec{b} and vec{c} are non-collinear vectors a | vedclass

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(D) Given $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$. Taking dot product with $\vec{c}$ on both sides: $(\vec{c} \cdot \vec{c}) (\vec{a} \cdot \vec{c

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$\cos 1$

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(D) Given $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$. Taking dot product with $\vec{c}$ on both sides: $(\vec{c} \cdot \vec{c}) (\vec{a} \cdot \vec{c}) = \vec{c} \cdot \vec{c} = |\vec{c}|^2$. Since $\vec{c}$ is non-zero, $|\vec{c}|^2 (\vec{a} \cdot \vec{c}) = |\vec{c}|^2$, so $\vec{a} \cdot \vec{c} = 1$ $(i)$. Given equation: $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 \beta - \sin \alpha) \vec{b} + (\beta^2 - 1) \vec{c}$. Rearranging terms: $(\vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{b}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (4 - 2 \beta - \sin \alpha) \vec{b} + (\beta^2 - 1) \vec{c}$. Since $\vec{b}$ and $\vec{c}$ are non-collinear, we compare coefficients: $\vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{b} = 4 - 2 \beta - \sin \alpha$ $(ii)$ $-\vec{a} \cdot \vec{b} = \beta^2 - 1 \Rightarrow \vec{a} \cdot \vec{b} = 1 - \beta^2$ $(iii)$. Substituting $(i)$ and $(iii)$ into $(ii)$: $1 + (1 - \beta^2) = 4 - 2 \beta - \sin \alpha$. $2 - \beta^2 = 4 - 2 \beta - \sin \alpha \Rightarrow \beta^2 - 2 \beta + 2 - \sin \alpha = 0$. For this to hold for any $\alpha, \beta$, we set $\sin \alpha = 1$ (i.e.,$\alpha = \frac{\pi}{2}$), then $\beta^2 - 2 \beta + 1 = 0 \Rightarrow (\beta - 1)^2 = 0 \Rightarrow \beta = 1$. Thus, $\sin (\alpha + \beta) = \sin (\frac{\pi}{2} + 1) = \cos 1$.

$\vec{b}$ and $\vec{c}$ are non-collinear vectors and $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$. If $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 \beta - \sin \alpha) \vec{b} + (\beta^2 - 1) \vec{c}$, then $\sin (\alpha + \beta) =$

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