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

$\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors. If $\alpha \vec{d}=\vec{a}+\vec{b}+\vec{c}$, $\beta \vec{a}=\vec{b}+\vec{c}+\vec{d}$, then $|\vec{a}+\vec{b}+\vec{c}+\vec{d}|=$
  1. $1$
  2. $2$
  3. $|\vec{a}-\vec{b}-\vec{c}|$
  4. $0$

Solution

$\vec{a}+\vec{b}+\vec{c}=\alpha \vec{d}$ [Given] $\Rightarrow \vec{b}+\vec{c}=\alpha \vec{d}-\vec{a}$ ....(i) $\vec{b}+\vec{c}+\vec{d}=\beta \vec{a}$ [Given] $\Rightarrow \vec{b}+\vec{c}=\beta \vec{a}-\vec{d}$ ....(ii) $\Rightarrow \alpha \vec{d}-a=\beta \vec{a}-\vec{d}$ [From (i) and (ii)] $\Rightarrow \alpha \vec{d}+\vec{d}=\beta \vec{a}+\vec{a} \Rightarrow(\alpha+1) \vec{d}=(\beta+1) \vec{a}$ ....(iii) Since, $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar $\Rightarrow \alpha=-1, \beta=-1$ $\therefore \vec{a}+\vec{b}+\vec{c}+\vec{d}=0$ [From (i)] $\Rightarrow|\vec{a}+\vec{b}+\vec{c}+\vec{d}|=0$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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