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Vector Algebra

Question
CBSEENMA12034223
Wired Faculty App

Find the area of the triangle formed by the points A (1, 1, 1), B (1, 2, 3) and C (2, 3, 1) with reference to a rectangular system of axes.

Solution
The given vertices are A (1, 1, 1), B (1, 2. 3) and C (2, 3, 1).
  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 be the position vectors of A, B, C respectively.
therefore 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 space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top comma space space straight c with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space straight k with hat on top
Now,       BC with rightwards arrow on top space equals space straight P. straight V. space of space straight C space minus space straight P. straight V. space of space straight B space equals space straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space minus 2 space straight k with hat on top
               BA 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 B space equals space straight a with rightwards arrow on top space space minus straight b with rightwards arrow on top space equals space minus straight j with hat on top space minus space 2 space straight k with hat on top
       BC with rightwards arrow on top space cross times space BA with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell space straight j with hat on top end cell cell space space space straight k with hat on top end cell row 1 cell space 1 end cell cell negative 2 end cell row 0 cell negative 1 end cell cell negative 2 end cell end table close vertical bar space equals space straight i with hat on top space open vertical bar table row cell space space 1 end cell cell space space minus 2 end cell row cell negative 1 end cell cell space space minus 2 end cell end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 1 cell negative 2 end cell row 0 cell negative 2 end cell end table close vertical bar space plus straight k with hat on top space open vertical bar table row 1 cell space space space 1 end cell row 0 cell negative 1 end cell end table close vertical bar space space space space space space space space space space space space space space space space equals space left parenthesis negative 2 minus 4 right parenthesis space straight i with hat on top space minus space left parenthesis negative 2 minus 0 right parenthesis space straight j with hat on top space plus space left parenthesis negative 1 minus 0 right parenthesis space straight k with hat on top space equals space minus 4 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space straight k with hat on top therefore space space space space open vertical bar BC with rightwards arrow on top space cross times space BA with rightwards arrow on top close vertical bar space equals space square root of 16 plus 4 plus 1 end root space equals space square root of 21 Area space of space increment ABC space equals space 1 half open vertical bar BC with rightwards arrow on top space cross times space BA with rightwards arrow on top close vertical bar space equals space 1 half square root of 21 space sq. space units.

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