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Vector Algebra

Question
CBSEENMA12034260

If 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 space and straight b with rightwards arrow on top space equals space straight j with hat on top space minus space straight k with hat on top.  Find a vector straight c with rightwards arrow on top such that straight a with rightwards arrow on top cross times space straight c with rightwards arrow on top space equals space straight b with rightwards arrow on top and straight a with rightwards arrow on top. space straight c with rightwards arrow on top space equals space 3.

Solution

Here ,    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 straight b with rightwards arrow on top space equals space straight j with hat on top space minus space straight k with hat on top
Let     straight c with rightwards arrow on top space equals space straight x space straight i with hat on top space plus space straight y space straight j with hat on top space plus space straight z space straight k with hat on top
Now,              straight a with rightwards arrow on top space cross times space straight c with rightwards arrow on top space equals space straight b with rightwards arrow on top
space rightwards double arrow space space space space space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 1 1 1 row straight x straight y straight z end table close vertical bar space equals straight j with hat on top space minus space straight k with hat on top
space rightwards double arrow space space space space straight i with hat on top space open vertical bar table row 1 1 row straight y straight z end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 1 1 row straight x straight z end table close vertical bar space plus space straight k with hat on top space open vertical bar table row 1 1 row straight x straight y end table close vertical bar space equals space straight j with hat on top space minus space straight k with hat on top
space rightwards double arrow space space left parenthesis straight z minus straight y right parenthesis space straight i with hat on top space minus space left parenthesis straight z minus straight x right parenthesis space straight j with hat on top space space plus space left parenthesis straight y minus straight x right parenthesis space straight k with hat on top space equals space straight j with hat on top space minus space straight k with hat on top space space space space
rightwards double arrow space space space space space space straight z minus straight y space equals space 0                                                         ...(1)
           x - z = 1                                                            ...(2)
            x - y = 1                                                           ...(3)
Also,     straight a with rightwards arrow on top. space straight c with rightwards arrow on top space equals 3 space space space space space space space space space space space space space space space space space rightwards double arrow space space space left parenthesis 1 right parenthesis thin space left parenthesis straight x right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis straight y right parenthesis space plus space left parenthesis 1 right parenthesis space left parenthesis straight z right parenthesis space equals space 3
therefore space space space space space space straight x plus straight y plus straight z space equals space 3                                                    ...(4)
From (1),    y = z
therefore    from (4), we get,
                       straight x plus 2 space straight y space equals space 3                                          ...(5)
Subtracting (3) from (5), we get,
                   3 straight y space equals space 2 space space space space or space space space straight y space equals space 2 over 3 space equals space straight z
therefore space space space from space left parenthesis 5 right parenthesis comma space space space straight x plus 4 over 3 space equals space 3 space space space space space rightwards double arrow space space space space straight x space equals space 3 space minus space 4 over 3 space equals space 5 over 3
therefore space space space we space have space straight x space equals space 5 over 3 comma space space space straight y space equals space space 2 over 3 comma space space space straight z space equals space 2 over 3
therefore space space space straight c with rightwards arrow on top space equals space 5 over 3 straight i with hat on top space space plus space 2 over 3 straight j with hat on top space plus space 2 over 3 straight k with hat on top