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Vector Algebra

Question
CBSEENMA12034018

If A (2, – 1, 3), B (8, 5, – 6) and C (4, 1, 0) are the vertices of a triangle. show that
(i) AB = 3 AC 
(ii) The direction-cosines of BC with rightwards arrow on top are fraction numerator negative 2 over denominator square root of 17 end fraction comma space fraction numerator negative 2 over denominator square root of 17 end fraction comma space fraction numerator 3 over denominator square root of 17 end fraction.

Solution
Let O be the origin.
A is (2, -1, 3)          rightwards double arrow             OA with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 3 space straight k with hat on top
B is (8, 5, -6)          rightwards double arrow             OB with rightwards arrow on top space equals space 8 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space space 6 space straight k with hat on top
C is (4, 1, 0)            rightwards double arrow              OC with rightwards arrow on top space equals space 4 space straight i with hat on top space plus space straight j with hat on top space plus space 0 space straight k with hat on top space equals space 4 space straight i with hat on top space plus space straight j with hat on top
(i)  AB with rightwards arrow on top space equals space OB with rightwards arrow on top space minus space OA with rightwards arrow on top space equals space open parentheses 8 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top close parentheses space minus space left parenthesis 2 space straight i with hat on top space minus space straight j with hat on top space plus space space 3 space straight k with hat on top right parenthesis space equals space 6 space straight i with hat on top space plus space 6 space straight j with hat on top space minus 9 space straight k with hat on top
therefore space space space space space AB space equals space square root of left parenthesis 6 right parenthesis squared plus left parenthesis 6 right parenthesis squared plus left parenthesis negative 9 right parenthesis squared end root space equals space square root of 36 plus 36 plus 81 end root
space space space space space space space space space space space space space space space equals space square root of 153 space equals square root of 9 space cross times space 17 end root space equals space 3 space square root of 17
     AC with rightwards arrow on top space equals space OC with rightwards arrow on top space minus space OA with rightwards arrow on top space equals space left parenthesis 4 space straight i with hat on top space plus space straight j with hat on top right parenthesis space minus space left parenthesis 2 space straight i with hat on top space minus space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis space equals space 2 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top
therefore space space space AC space equals space square root of left parenthesis 2 right parenthesis squared plus left parenthesis 2 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared end root space equals space square root of 4 plus 4 plus 9 end root space equals space square root of 17
therefore space space AB space equals space 3 space space AC
(ii)  BC with rightwards arrow on top space equals space OC with rightwards arrow on top space minus space OB with rightwards arrow on top space equals space left parenthesis 4 space straight i with hat on top space plus space straight j with hat on top right parenthesis space minus space left parenthesis 8 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 6 space straight k with hat on top right parenthesis space equals space minus 4 space straight i with hat on top space minus space 4 space straight j with hat on top space plus space 6 space straight k with hat on top
therefore space space space BC space equals space square root of left parenthesis negative 4 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus left parenthesis 6 right parenthesis squared end root space equals square root of 16 plus 16 plus 36 end root space equals square root of 68 equals space square root of 4 space cross times 17 end root space equals space 2 space square root of 17
therefore space space space space space straight a space unit space vector space along space BC with rightwards arrow on top space equals space fraction numerator BC with rightwards arrow on top over denominator open vertical bar BC with rightwards arrow on top close vertical bar end fraction space equals fraction numerator BC with rightwards arrow on top over denominator BC end fraction
                                 equals space fraction numerator 1 over denominator 2 square root of 17 end fraction left parenthesis negative 4 space straight i with hat on top space minus space 4 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis space equals space minus fraction numerator 2 over denominator square root of 17 end fraction straight i with hat on top space minus space fraction numerator 2 over denominator square root of 17 end fraction straight j with hat on top space plus space fraction numerator 3 over denominator square root of 17 end fraction straight k with hat on top
therefore space space space direction cosines of BC with rightwards arrow on top i.e.,   coeffs. of straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top  are negative fraction numerator 2 over denominator square root of 17 end fraction comma space minus fraction numerator 2 over denominator square root of 17 end fraction comma space fraction numerator 3 over denominator square root of 17 end fraction.