Question

If $straight a with rightwards arrow on top$ makes equal angles with $straight i with hat on top comma space straight j with hat on top space and space straight k with hat on top$ and has magnitude 3, then prove that the angle between $straight a with rightwards arrow on top$ and each of $straight i with hat on top comma space straight j with hat on top space and space straight k with hat on top$ is $cos to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses.$

Solution

Let $straight a with rightwards arrow on top space equals straight a subscript 1 straight i with hat on top space plus space straight a subscript 2 straight j with hat on top space plus space straight a subscript 3 straight k with hat on top$ such that $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 3$ and O be the angle which $straight a with rightwards arrow on top$ makes with $straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top.$
$therefore space space space space cos space straight theta space equals space fraction numerator straight a with rightwards arrow on top. space straight i with hat on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar. space open vertical bar straight i with hat on top close vertical bar end fraction space equals space fraction numerator straight a subscript 1 over denominator left parenthesis 3 right parenthesis left parenthesis 1 right parenthesis end fraction space space space rightwards double arrow space space space space straight a subscript 1 space equals space 3 space cos space straight theta$
Again $cos space straight theta space equals space fraction numerator straight a with rightwards arrow on top. space straight j with hat on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar. space open vertical bar straight j with hat on top close vertical bar end fraction space equals space fraction numerator straight a subscript 2 over denominator left parenthesis 3 right parenthesis thin space left parenthesis 1 right parenthesis end fraction space space space rightwards double arrow space space straight a subscript 2 space equals space 3 space cos space straight theta$
and $cos space straight theta space equals space fraction numerator straight a with rightwards arrow on top. space straight k with hat on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar. space open vertical bar straight j with hat on top close vertical bar end fraction space equals space fraction numerator straight a subscript 3 over denominator left parenthesis 3 right parenthesis thin space left parenthesis 1 right parenthesis space end fraction space space rightwards double arrow space space straight a subscript 3 space equals space 3 space cos space straight theta$
Also, $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 3$ $rightwards double arrow space space space space space square root of straight a subscript 1 squared plus straight a subscript 2 squared plus straight a subscript 3 squared end root space equals space 3$
$rightwards double arrow space space space square root of 9 space cos squared straight theta space plus space 9 space cos squared straight theta space plus space 9 space cos squared space straight theta end root space equals space 3 space space space space space space space rightwards double arrow space space space square root of 27 space cos squared straight theta end root space equals space 3 rightwards double arrow space space space 3 square root of 3 space cos space straight theta space equals space 3 space space space space space space rightwards double arrow space space space cos space straight theta space equals space fraction numerator 1 over denominator square root of 3 end fraction therefore space space space space space space space straight theta space space equals space cos to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses$

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Vector Algebra

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