If the vertices of a triangle are $A (2, - 1, 1)$ , $B (1, - 3, - 5)$ and $C (3, 4, - 4)$ , then prove by vector method that it is a right-angled triangle.

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Solution

$\to A = 2^I -^J +^k \to B =^I - 3^J - 5^k \to C = 3^I + 4^J - 4^k - - \to A B = \to B - \to A = (^I - 3^J - 5^k) - (2^I -^J +^k) = -^I - 2^J - 6^k - - \to B C = \to C - \to B = (3^I + 4^J - 4^k) - (^I - 3^J - 5^k) = 2^I + 7^J +^k - - \to A C = \to C - \to A = (3^I + 4^J - 4^k) - (2^I -^J +^k) =^I + 5^J - 5^k t w o v e c t o r s a r e p r p e n d i c u l a r t o e a c h o t h e r, i f t h e i r s c a l a r p r o d u c t i s z e r o - - \to A B . - - \to B C = (-^I - 2^J - 6^k) . (2^I + 7^J +^k) = - 2 - 14 - 6 = - 22 - - \to B C . - - \to A C \neq 0 - - \to A B . - - \to A C \neq 0 n o 2 v e c t o r s a r e p r e p e n d i c u l a r t o e a c h o t h e r \to c o o r d i n a t e s o f C m u s t b e (3, - 4, - 4) t o p r o v e t h e t r i a n g l e i s r i g h t a n g l e d .$

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