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Vector Algebra

Question
CBSEENMA12034098
Wired Faculty App

If straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top comma space 2 space straight i with hat on top plus 5 space straight j with hat on top comma space space 3 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 space space and space straight i with hat on top space minus space 6 space straight j with hat on top space minus space straight k with hat on top, respectively, are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Deduce that AB and CD are parallel.

Solution

Let straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top comma space straight d with rightwards arrow on top be position vectors of points A, B, C, D respectively.
therefore space space space space straight a with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space 5 space straight j with hat on top comma space straight c with rightwards arrow on top space equals space 3 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 comma space straight d with rightwards arrow on top space equals space straight i with hat on top space minus space 6 space straight j with hat on top space minus space straight k with hat on top
   AB with rightwards arrow on top space equals space straight P. straight V. space of space straight B space minus space space straight P. straight V. space of space straight A space equals space straight b with rightwards arrow on top minus straight a with rightwards arrow on top space equals space straight i with hat on top plus 4 space straight j with hat on top space minus space straight k with hat on top CD with rightwards arrow on top space equals space straight P. straight V. space of space straight D space minus space straight P. straight V. space of space straight C space equals straight d with rightwards arrow on top space minus space straight c with rightwards arrow on top space equals space minus 2 space straight i with hat on top space minus space 8 space straight j with hat on top space plus space 2 space straight k with hat on top
therefore space space space open vertical bar AB with rightwards arrow on top close vertical bar space equals space square root of 1 plus 16 plus 1 end root space equals space square root of 18 comma space space space open vertical bar CD with rightwards arrow on top close vertical bar space equals space square root of 4 plus 64 plus 4 end root space equals square root of 72
         AB with rightwards arrow on top space. space CD with rightwards arrow on top space equals space left parenthesis 1 right parenthesis thin space left parenthesis negative 2 right parenthesis space plus space left parenthesis 4 right parenthesis thin space left parenthesis negative 8 right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis 2 right parenthesis space equals space minus 2 minus 32 minus 2 space equals space minus 36
Let straight theta be angle between AB and CD
therefore space space cos space straight theta space equals space fraction numerator AB with rightwards arrow on top. space CD with rightwards arrow on top over denominator open vertical bar AB with rightwards arrow on top close vertical bar space space open vertical bar CD with rightwards arrow on top close vertical bar end fraction space equals space fraction numerator negative 36 over denominator left parenthesis square root of 18 right parenthesis space left parenthesis square root of 72 right parenthesis end fraction space equals space fraction numerator negative 36 over denominator 3 square root of 2 space cross times space 6 space square root of 2 end fraction space equals negative 36 over 36 space equals space minus 1 therefore space space space straight theta space equals space straight pi space as space 0 space less or equal than space straight theta space less or equal than space straight pi therefore space space AB space and space CD space are space parallel.

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