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Vector Algebra

Question
CBSEENMA12034123
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If a unit vector straight a with rightwards arrow on top makes angles straight pi over 3 with straight i with hat on top comma  straight pi over 4 space with space straight j with hat on top and acute angle straight theta with straight k with hat on top, then find the components of straight a with rightwards arrow on top and the angle straight theta.

Solution

Let straight a with rightwards arrow on top space equals space straight a subscript 1 straight i with hat on top space plus space straight a subscript 2 straight j with hat on top space plus space straight a subscript 3 straight k with hat on top
  Since straight a with rightwards arrow on top makes an angle straight pi over 3 with straight i with hat on top
therefore space space space cos space straight pi over 3 space equals space fraction numerator straight a with rightwards arrow on top. space straight i with hat on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar. space open vertical bar straight i with hat on top close vertical bar end fraction space space space space rightwards double arrow space space space 1 half space equals space fraction numerator straight a subscript 1 over denominator left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis end fraction space space space space space space open square brackets because space straight a with rightwards arrow on top space is space straight a space unit space vector close square brackets rightwards double arrow space space space space space straight a subscript 1 space equals space 1 half
Again straight a with rightwards arrow on top makes an angle straight pi over 4 space with space straight j with hat on top
therefore space space cos space straight pi over 4 space equals space fraction numerator straight a with rightwards arrow on top. space straight j with hat on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar. space open vertical bar straight j with hat on top close vertical bar end fraction space space space rightwards double arrow space space space space fraction numerator 1 over denominator square root of 2 end fraction space equals space fraction numerator straight a subscript 2 over denominator left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis end fraction space space rightwards double arrow space space space space straight a subscript 2 space equals space fraction numerator 1 over denominator square root of 2 end fraction
Since straight a with rightwards arrow on top is a unit vector
therefore space space space space straight a subscript 1 squared plus straight a subscript 2 squared plus straight a subscript 3 squared space equals space 1
therefore space space space 1 fourth plus 1 half plus straight a subscript 3 squared space equals space 1 space space space space space space space space space space space space space space space space space space open square brackets because space space straight a subscript 1 space equals space 1 half comma space straight a subscript 2 space equals fraction numerator 1 over denominator square root of 2 end fraction close square brackets rightwards double arrow space space space space straight a subscript 3 squared space equals space 1 fourth space space space space space rightwards double arrow space space space space straight a subscript 3 space equals plus-or-minus 1 half
Now straight theta is the angle between straight a with rightwards arrow on top space and space straight k with hat on top
therefore space space space cos space straight theta space equals space fraction numerator straight a with rightwards arrow on top. space straight k with hat on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar. space stack open vertical bar straight k close vertical bar with hat on top end fraction space equals space fraction numerator straight a subscript 3 over denominator left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis end fraction space equals plus-or-minus 1 half therefore space space space space straight theta space equals space straight pi over 3 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 space space space space space space space space space space space space open square brackets because space 0 space less or equal than space straight theta space less or equal than space straight pi space and space straight theta space is space acute close square brackets

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