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Vector Algebra

Question
CBSEENMA12034128
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Dot product of a vector with vectors 3 space straight i with hat on top minus 5 space straight k with hat on top comma space space space 2 straight i with hat on top plus space 7 space straight j with hat on top space and space straight i with hat on top plus straight j with hat on top plus straight k with hat on top are respectively – 1, 6 and 5. Find the vector.

Solution

Let straight r with rightwards arrow on top space equals space straight x straight i with hat on top space plus space straight y straight j with hat on top plus straight z straight k with hat on top be required vector.
From given conditions,
            left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis. space left parenthesis 3 straight i with hat on top space minus 5 straight k with hat on top right parenthesis space equals space minus 1 space space space space space space space rightwards double arrow space space space 3 straight x minus 5 straight z space equals space minus 1         ...(1)
            left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis. space left parenthesis 2 straight i with hat on top plus 7 straight j with hat on top right parenthesis space equals space 6 space space space rightwards double arrow space space space 2 straight x plus 7 straight y space equals space 6                       ...(2)
            left parenthesis straight x straight i with hat on top plus straight y straight j with hat on top plus straight z straight k with hat on top right parenthesis. space left parenthesis straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis space equals space 5 space space space rightwards double arrow space space space straight x plus straight y plus straight z space equals space 5                     ...(3)
        7 space cross times space left parenthesis 3 right parenthesis space minus space left parenthesis 2 right parenthesis space gives comma space       5 straight x plus 7 straight z space equals 29                                          ...(4)
Multiplying (2) by 7 and (4) by 5, we get,
         21 x - 35 z = -7 
          25 x+35 z = 145
Adding these equation, 46x = 138  rightwards double arrow space space straight x space equals space 3
therefore space space from space left parenthesis 2 right parenthesis comma space 6 plus 7 straight y space equals space 6 space space space space rightwards double arrow space space space 7 straight y space equals space 0 space space space space rightwards double arrow space space straight y space equals space 0 From space left parenthesis 1 right parenthesis comma space space 9 space minus space 5 straight z space equals space minus 1 space space space space rightwards double arrow space space space minus 5 straight z space equals space minus 10 space rightwards double arrow space straight z space equals space 2 therefore space space space space straight r with rightwards arrow on top space equals space 3 straight i with hat on top plus 2 straight k with hat on top

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