Let , , be distinct real numbers. The points with position vectors i + j + k , i + j + k and i + j + k

Step-by-step solution

, and be distinct real numbers aligned & a = i + j + k & b = i + j + k & c = i + j + k aligned Let =1, =2 and =3 then, a = i +2 j +3 k aligned & b =2 i +3 j + k & c =3 i +2 j +2 k aligned Now, | ab |= (2-1)^2 +(3-2)^2+(1-3)^2= 6 Similarly | bc |=| ac |= 6 Hence, point for equilateral triangle.