Let a , b anc c be distinct and real numbers . The points with position vectors ai+bj+ck , bi+cj+ak and ci+aj+bk Then

Answers (1)

Solution:   Let the given  position vectors  be of points 

         A , B and C respectively , then 

           \ \ Rightarrow hspace1cmleft | vecAB 
ight |=sqrt(b-a)^2+(c-b)^2+(a-c)^2\ \ Rightarrow hspace1cmleft | vecBC 
ight |=sqrt(c-b)^2+(a-c)^2+(a-b)^2\ \ Rightarrow hspace1cmleft | vecCA 
ight |=sqrt(a-c)^2+(b-a)^2+(c-b)^2\ \ 	herefore hspace1cmleft | vecAB 
ight |=left | vecBC 
ight |=left | vecCA 
ight |

Hence ,    igtriangleup ABC is an  equilateral triangle 

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