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Vector Algebra

Question
CBSEENMA12034121

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

Solution
Let a1 , a, a3 be scalar components of straight a with rightwards arrow on top.
   space therefore space space space space straight a with rightwards arrow on top space equals space straight a subscript 1 space straight i with hat on top space plus space straight a subscript 2 space straight j with hat on top space plus space straight a subscript 3 space straight k with hat on top
 Since straight a with rightwards arrow on top makes an angle straight pi over 4 space with space straight i 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 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 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 1 over denominator left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis end fraction space space space space space space space space space space space space space open square brackets because space space space open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 1 close square brackets
rightwards double arrow space space space straight a subscript 1 space equals space fraction numerator 1 over denominator square root of 2 end fraction
Again, straight a with rightwards arrow on top makes an angle straight pi over 3 space with space straight j with hat on top
therefore 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 j with hat on top close vertical bar end fraction space space space space space rightwards double arrow space space space space 1 half space equals space fraction numerator straight a subscript 2 over denominator left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis space end fraction space space space space space rightwards double arrow space space space straight a subscript 2 space equals space 1 half
Since straight a with rightwards arrow on top is a unit vector
therefore space space open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 1 space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared space equals space 1 space space space space space rightwards double arrow 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 half plus 1 fourth plus straight a subscript 3 squared space equals space 1 space space space space space space rightwards double arrow space space space straight a subscript 3 squared space equals 1 fourth space space space space rightwards double arrow space straight a subscript 3 space equals space plus-or-minus 1 half
Since space straight theta space space is space angle space between space straight a with rightwards arrow on top space and space straight k with hat on top
therefore 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 open vertical bar straight k with hat on top close vertical bar end fraction space equals fraction numerator straight a subscript 3 over denominator left parenthesis 1 right parenthesis left parenthesis 1 right parenthesis end fraction space equals space plus-or-minus 1 half.

∴    = 60°, 120°
Rejecting straight theta = 120° as ө is given to be acute,
we have straight theta = 160°
Now scalar components of straight a with rightwards arrow on top are fraction numerator 1 over denominator square root of 2 end fraction comma space 1 half comma space plus-or-minus 1 half and vector component of straight a with rightwards arrow on top are fraction numerator 1 over denominator square root of 2 end fraction space straight i with hat on top comma space 1 half space straight j with hat on top space plus-or-minus 1 half straight k with hat on top.
Also   straight theta space equals space 60 degree.