Let `alpha, beta, gamma` be distinct real numbers. The points with position vectors `alpha hati + beta hatj +gamma hat k , beta hati + gamma hatj +alpha hat k , gamma hati +alpha hatj + beta hatk`

Options

A. are collinear
B. form an equilateral triangle
C. form a scalene triangle
D. form a right-angled triangle

Answer

B

Explanation

Let the given position vectors be of points A, B and C, respectively. Then
`|vec(AB)| = sqrt((beta -alpha)^(2)+ (gamma-beta)^(2) + (alpha -gamma)^(2))`
`|vec(BC)| = sqrt((gamma-beta)^(2)+ (alpha -gamma)^(2) + (alpha -beta)^(2))`
`|vec(CA)| = sqrt((alpha-gamma)^(2)+ (beta-alpha)^(2)+ (gamma-beta)^(2))`
`because |vec(AB)| = |vec(BC)| = |vec(CA)|`
Hence, `DeltaABC` is an equilateral triangle.