vec{a}, vec{b}, vec{c} are non-coplanar vecto | vedclass

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(D) Given: $\vec{a}+\vec{b}+\vec{c}=\alpha \vec{d}$  ....$(i)$ Given: $\vec{b}+\vec{c}+\vec{d}=\beta \vec{a}$  ....(ii) From $(i)$,$\vec{b}+\vec{c}=\a

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(D) Given: $\vec{a}+\vec{b}+\vec{c}=\alpha \vec{d}$  ....$(i)$ Given: $\vec{b}+\vec{c}+\vec{d}=\beta \vec{a}$  ....(ii) From $(i)$,$\vec{b}+\vec{c}=\alpha \vec{d}-\vec{a}$. Substituting this into (ii): $(\alpha \vec{d}-\vec{a})+\vec{d}=\beta \vec{a}$. Rearranging the terms: $(\alpha+1)\vec{d} = (\beta+1)\vec{a}$. Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$\vec{d}$ must be a linear combination of $\vec{a}, \vec{b}, \vec{c}$. For the equation $(\alpha+1)\vec{d} = (\beta+1)\vec{a}$ to hold for non-coplanar vectors,the coefficients must be zero,implying $\alpha+1=0$ and $\beta+1=0$,so $\alpha=-1$ and $\beta=-1$. Substituting $\alpha=-1$ into $(i)$: $\vec{a}+\vec{b}+\vec{c} = -\vec{d}$. Therefore,$\vec{a}+\vec{b}+\vec{c}+\vec{d} = \vec{0}$. Thus,$|\vec{a}+\vec{b}+\vec{c}+\vec{d}| = |\vec{0}| = 0$.

$\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors. If $\alpha \vec{d}=\vec{a}+\vec{b}+\vec{c}$ and $\beta \vec{a}=\vec{b}+\vec{c}+\vec{d}$,then $|\vec{a}+\vec{b}+\vec{c}+\vec{d}|=$

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