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Vector Algebra

Question
CBSEENMA12034122
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If A, B, C have position vectors (0, 1, 1) (3, 1, 5), (0. 3, 3) respectively, then show that the ∆ABC is right angled at C.

Solution

Let straight a with rightwards arrow on top space equals space left parenthesis 0 comma space 1 comma space 1 right parenthesis space equals space straight j with hat on top space plus space straight k with hat on top comma space space space space space space straight b with rightwards arrow on top space equals space left parenthesis 3 comma space 1 comma space 5 right parenthesis space equals space 3 space straight i with hat on top space plus space straight j with hat on top space plus space 5 space straight k with hat on top comma
      straight i with hat on top space equals space left parenthesis 0 comma space 3 comma space 3 right parenthesis space equals space 3 straight j with hat on top plus 3 straight k with hat on top be the position vectors of A, B, C respectively.
    CA with rightwards arrow on top space equals space straight P. straight V. space of space straight A space minus space straight P. straight V. space of space straight C space equals space straight a with rightwards arrow on top space minus space straight c with rightwards arrow on top space equals space minus 2 space straight j with hat on top space minus space 2 space straight k with hat on top CB with rightwards arrow on top space equals space straight P. straight V. space of space straight B space minus space straight P. straight V. space of space straight C space equals space straight b with rightwards arrow on top minus straight c with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top CA with rightwards arrow on top. space CB with rightwards arrow on top space equals space left parenthesis negative 2 straight j with hat on top space minus space 2 straight k with hat on top right parenthesis. space left parenthesis 3 straight i with hat on top space minus space 2 straight j with hat on top plus 2 straight k with hat on top right parenthesis space space space space space space space space space space space space space space space space equals space left parenthesis 0 right parenthesis thin space left parenthesis 3 right parenthesis space plus space left parenthesis negative 2 right parenthesis space left parenthesis negative 2 right parenthesis plus space left parenthesis negative 2 right parenthesis thin space left parenthesis 2 right parenthesis space equals space 0 plus 4 minus 4 space equals space 0 therefore space space space CA with rightwards arrow on top space and space CB with rightwards arrow on top space are space perpendicular therefore space space space space increment ABC space is space right space angled space at space straight C.

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