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If $a(\alpha \times \beta)+b(\beta \times \gamma)+c(\gamma \times \alpha)=0$ and atleast one of the scalars $a, b, c$ is non-zero, then the vectors $\alpha, \beta, \gamma$ are
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Verified Answer
The correct answer is:
coplanar
Given that,
$a(\alpha \times \beta)+b(\beta \times \gamma)+c(\gamma \times \alpha)=0$
On taking dot product of $\alpha$ with the given cross product, we get
$\begin{aligned}
& a \alpha \cdot(\alpha \times \beta)+b \alpha \cdot(\beta \times \gamma)+c \alpha \cdot(\gamma \times \alpha)=0 \\
& a[\alpha \alpha \beta]+b(\alpha \beta \gamma)+c(\alpha \gamma \alpha)=0 \\
& \therefore \quad[x x y]=[x y x]=[y x x]=0 \\
& \Rightarrow \quad b[\alpha \alpha \beta]=0
\end{aligned}$
Since, $\quad b \neq 0$
$\therefore \quad[\alpha \alpha \beta]=0$
Hence, $\alpha, \beta, \gamma$ are coplanar.
$a(\alpha \times \beta)+b(\beta \times \gamma)+c(\gamma \times \alpha)=0$
On taking dot product of $\alpha$ with the given cross product, we get
$\begin{aligned}
& a \alpha \cdot(\alpha \times \beta)+b \alpha \cdot(\beta \times \gamma)+c \alpha \cdot(\gamma \times \alpha)=0 \\
& a[\alpha \alpha \beta]+b(\alpha \beta \gamma)+c(\alpha \gamma \alpha)=0 \\
& \therefore \quad[x x y]=[x y x]=[y x x]=0 \\
& \Rightarrow \quad b[\alpha \alpha \beta]=0
\end{aligned}$
Since, $\quad b \neq 0$
$\therefore \quad[\alpha \alpha \beta]=0$
Hence, $\alpha, \beta, \gamma$ are coplanar.
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