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Vector Algebra

Question
CBSEENMA12034225
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Find the area of the triangle with vertices (1, 1, 2), (2, 3, 5) and (1, 5, 5). 

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 be position vectors of A, B, C repsectively.
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 2 space straight k with hat on top comma space space space space straight b 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 5 space straight k with hat on top comma space space straight c with rightwards arrow on top space equals straight i with hat on top space plus space 5 space straight j with hat on top space plus space 5 space straight k with hat on top
Now, BC with rightwards arrow on top space equals straight P. straight V. space of space straight C space minus space straight P. straight V. space of space straight B space equals straight c with rightwards arrow on top minus straight b with rightwards arrow on top space equals space minus straight i with hat on top space plus space 2 space straight j 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 thin space straight B space equals space straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top space equals space minus straight i with hat on top space minus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top
therefore space space space space 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 space straight j with hat on top end cell cell space space straight k with hat on top end cell row cell negative 1 end cell cell space space space space 2 end cell cell space space 0 end cell row cell negative 1 end cell cell space minus 2 end cell cell negative 3 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 space 2 end cell 0 row cell negative 2 end cell cell negative 3 end cell end table close vertical bar minus space straight j with hat on top space open vertical bar table row cell negative 1 end cell 0 row cell negative 1 end cell cell negative 3 end cell end table close vertical bar plus space straight k with hat on top space open vertical bar table row cell negative 1 end cell 2 row cell negative 1 end cell cell negative 2 end cell end table close vertical bar space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space left parenthesis negative 6 minus 0 right parenthesis space straight i with hat on top space minus space left parenthesis 3 minus 0 right parenthesis space straight j with hat on top space plus space left parenthesis 2 plus 2 right parenthesis space straight k with hat on top space equals space minus 6 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top space space space space space space space space space space 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 36 plus 9 plus 16 end root space equals space square root of 61
Area of 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 61 space sq. space units

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