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If the vectors $\alpha \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}, \hat{\mathrm{i}}+\hat{\mathrm{k}}$ and $\gamma \hat{\mathrm{i}}+\hat{\mathrm{j}}+\beta \hat{\mathrm{k}}$ lie on a
plane, where $\alpha, \beta$ and $\gamma$ are distinct non-negative numbers, then $\gamma$ is
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plane, where $\alpha, \beta$ and $\gamma$ are distinct non-negative numbers, then $\gamma$ is
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Verified Answer
The correct answer is:
Geometric mean of $\alpha$ and $\beta$
If three vectors are co-planar.
$\Rightarrow\left|\begin{array}{lll}\alpha & \alpha & \gamma \\ 1 & 0 & 1 \\ \gamma & \gamma & \beta\end{array}\right|=0$
$\Rightarrow \quad \alpha[0-\gamma]-\alpha[\beta-\gamma]+\gamma[\gamma-0]=0$
$\Rightarrow-\alpha \gamma-\alpha \beta+\alpha \gamma+\gamma^{2}=0$
$\Rightarrow \quad \gamma^{2}=\alpha \beta$
$\Rightarrow \quad$ So $\alpha, \gamma, \beta$ are in G.P.
$\Rightarrow\left|\begin{array}{lll}\alpha & \alpha & \gamma \\ 1 & 0 & 1 \\ \gamma & \gamma & \beta\end{array}\right|=0$
$\Rightarrow \quad \alpha[0-\gamma]-\alpha[\beta-\gamma]+\gamma[\gamma-0]=0$
$\Rightarrow-\alpha \gamma-\alpha \beta+\alpha \gamma+\gamma^{2}=0$
$\Rightarrow \quad \gamma^{2}=\alpha \beta$
$\Rightarrow \quad$ So $\alpha, \gamma, \beta$ are in G.P.
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