Let `alpha, beta , lambda` be distinct real numbers. The points with position vectros `alpha hat (i) + beta hat(j) + lambda hat(k), beta hat(i) + lambda hat(j) + alpha hat(k) and lambda hat(i) + alpha hat(j) + beta hat (k)`

Options

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

Answer

B

Explanation

`alpha, beta` and `gamma` be distinct real numbers
`vec(c)=alpha hat(i)+beta(j)+gammahat(k)`
`vec(c)=betahat(i)+gammahat(j)+alphahat(k)`
`vec(c)=gamma hat(i)+alpha hat(j)+beta(k)`
Let `alpha=1, beta=2 and gamma=3`
then, `vec(a)=hat(i)+3hat(j)+3hat(k)`
`vec(b)=2hat(i)+3hat(j)+hat(k)`
`vec(c)=3hat(i)+2hat(j)+2hat(k)`
Now, `|vec(ab)|=sqrt((2-1)^(2))+(3-2)^(2)+(1-3)^(2)=sqrt(6)`
Similarly `|vec(bc)|=|vec(ac)|=sqrt(6)`
Hence, point for equilateral triangle.