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Vector Algebra

Question
CBSEENMA12033985
Wired Faculty App

Find the direction cosines of the vector joining the points A (1, 2, – 3) and B (– 1, – 2. 1). directed from A to B.

Solution

A(1, 2, 3), B(-1, 2, 1) are given points. 
therefore space space space space space AB 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 A
                    equals left parenthesis negative straight i with hat on top space minus space 2 space straight j with hat on top space plus space straight k with hat on top right parenthesis space minus space left parenthesis straight i with hat on top space plus space 2 space straight i with hat on top space minus space 3 straight k with hat on top right parenthesis
therefore space space space space space space space open vertical bar AB with rightwards arrow on top close vertical bar space equals space minus 2 straight i with hat on top space minus space 4 straight j with hat on top space plus space 4 straight k with hat on top therefore space space space space space space open vertical bar AB with rightwards arrow on top close vertical bar space equals space minus 2 straight i with hat on top space minus space 4 space straight j with hat on top space plus space 4 space straight k with hat on top therefore space space space space space open vertical bar AB with rightwards arrow on top close vertical bar space equals space square root of left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared plus left parenthesis 4 right parenthesis squared end root space equals space square root of 4 plus 16 plus 16 end root space equals square root of 36 space equals 6 therefore space space space space unit space vector space along space AB with rightwards arrow on top space space space space space space space space space space space space space space space space space space equals space fraction numerator AB with rightwards arrow on top over denominator open vertical bar AB with rightwards arrow on top close vertical bar end fraction space equals 1 over 6 left parenthesis negative 2 space straight i with hat on top space minus space 4 space straight j with hat on top space plus space 4 space straight k with hat on top right parenthesis space space space space space space space space space space space space space space space space space space equals space minus 1 third straight i with hat on top space minus space 2 over 3 straight j with hat on top space plus space 2 over 3 straight k with hat on top therefore space space space space space required space direction space cosines space are space space minus 1 third comma space minus 2 over 3 comma space 2 over 3.

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