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  • 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

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

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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 

Posted by

Deependra Verma

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