Question

Let $\alpha ,\beta ,\gamma$ be three distinct real numbers. The points with position vectors are $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k},\beta \hat{i}+\gamma \hat{j}+\alpha \hat{k}\&\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}$

• A. Are collinear
• B. Form an equilateral triangle
• C. Form a scalene triangle
• D. Form a right angled triangle

Answer 1

The correct option is B

Form an equilateral triangle

Explanation For The Correct Option:

Finding the position of points with position vectors :

Considering the points

$$\begin{gathered} A=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k} \\ B=\beta \hat{i}+\gamma \hat{j}+\alpha \hat{k} \\ C=\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k} \end{gathered}$$

Therefore,

$$\begin{gathered} \overrightarrow{AB}=(\beta –\alpha )\hat{i}+(\gamma –\beta )\hat{j}+(\alpha –\gamma )\hat{k}\dots \dots \dots \left(i\right) \\ \overrightarrow{BC}=(\gamma –\beta )\hat{i}+(\alpha –\gamma )\hat{j}+(\beta –\alpha )\hat{k}\dots \dots \dots .\left(ii\right) \\ \overrightarrow{CA}=(\alpha –\gamma )\hat{i}+(\beta –\alpha )\hat{j}+(\gamma –\beta )\hat{k}\dots \dots \dots .\left(iii\right) \end{gathered}$$

Since we know the magnitude of a vector $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$

$\Rightarrow \left|\overrightarrow{a}\right|=\sqrt{x^2+y^2+z^2}$

Therefore,

$\begin{array}{l}\begin{array}{c}\left|\overrightarrow{AB}\right|=\sqrt{(\beta -\alpha {)}^2+(\gamma -\beta {)}^2+(\alpha -\gamma {)}^2}\\ \left|\overrightarrow{BC}\right|=\sqrt{(\beta -\alpha {)}^2+(\gamma -\beta {)}^2+(\alpha -\gamma {)}^2}\\ \left|\overrightarrow{CA}\right|=\sqrt{(\beta -\alpha {)}^2+(\gamma -\beta {)}^2+(\alpha -\gamma {)}^2}\end{array}\end{array}$

Thus, $\left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|=\left|\overrightarrow{CA}\right|$

As all the sides are equal in magnitude. Hence, it forms an equilateral triangle.

Hence, option (B) is correct.