If \(\alpha, \beta, \gamma\) are distinct real numbers and \(\alpha+\beta+\gamma \neq 0\), then the points…
If \(\alpha, \beta, \gamma\) are distinct real numbers and \(\alpha+\beta+\gamma \neq 0\), then the points with position vectors \(\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}}, \beta \hat{\mathbf{i}}+\gamma \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}}\) and \(\gamma \hat{\mathbf{i}}+\alpha \hat{\mathbf{j}}+\beta \hat{\mathbf{k}}\) are
collinear
vertices of a scalene triangle
vertices of an isosceles triangle
vertices of an equilateral triangle
Solution
Let position vectors are
\(\begin{aligned}
& \mathrm{O A}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}} \\
& \mathrm{O B}=\beta \hat{\mathrm{i}}+\gamma \hat{\mathrm{j}}+\alpha \hat{\mathrm{k}} \text {, and } \\
& \mathrm{O C}=\gamma \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}
\end{aligned}\)
So, \(\quad \mathrm{A B}=(\boldsymbol{\beta}-\boldsymbol{\alpha}) \hat{\mathrm{i}}+(\gamma-\boldsymbol{\beta}) \hat{\mathrm{j}}+(\boldsymbol{\alpha}-\gamma) \hat{\mathrm{k}}\)
\(\mathrm{B C}=(\gamma-\beta) \hat{\mathrm{i}}+(\alpha-\gamma) \hat{\mathrm{j}}+(\beta-\alpha) \hat{\mathrm{k}}\) and
\(\mathrm{C A}=(\alpha-\gamma) \hat{\mathrm{i}}+(\boldsymbol{\beta}-\alpha) \hat{\mathrm{j}}+(\gamma-\beta) \hat{\mathrm{k}}\)
\(\because|\mathrm{A B}|=\sqrt{(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2}\)
\(|\mathrm{B C}|=\sqrt{(\beta-\gamma)^2+(\gamma-\alpha)^2+(\alpha-\beta)^2}\)
and \(|\mathrm{C A}|=\sqrt{(\gamma-\alpha)^2+(\alpha-\beta)^2+(\beta-\gamma)^2}\)
\(\begin{aligned}
\because \quad|\mathbf{A B}| & =|\mathbf{B Q}|=|\mathbf{C A}| \\
& =\sqrt{(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2} \neq 0 \\
& {[\because \alpha, \beta, \gamma \text { are distinct real numbers }] }
\end{aligned}\)
So, given position vectors are vertices of an equilateral triangle.
Hence, option (d) is correct.