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Vector Algebra

Question
CBSEENMA12034263
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straight A with rightwards arrow on top comma space straight B with rightwards arrow on top comma space straight C with rightwards arrow on top are unit vectors,  Suppose that  straight A with rightwards arrow on top. space straight B with rightwards arrow on top space equals space straight A with rightwards arrow on top. space straight C with rightwards arrow on top space equals space 0 and the angle between B and C is straight pi over 6. Prove that straight A with rightwards arrow on top space equals plus-or-minus 2 space open parentheses straight B with rightwards arrow on top space cross times space straight C with rightwards arrow on top close parentheses.

Solution

Since  straight A with rightwards arrow on top. space straight B with rightwards arrow on top space equals space straight A with rightwards arrow on top. space straight C with rightwards arrow on top space equals 0
therefore space space space space straight A with rightwards arrow on top space is space perpendicular space to space straight B with rightwards arrow on top space as space well space as space straight C with rightwards arrow on top therefore space space space straight A with rightwards arrow on top space is space parallel space to space straight B with rightwards arrow on top space cross times space straight C with rightwards arrow on top
Let straight A with rightwards arrow on top space equals space straight lambda open parentheses straight B with rightwards arrow on top space cross times space straight C with rightwards arrow on top close parentheses                                                        ...(1)
where straight lambda is some scalar
therefore space space space open vertical bar straight A with rightwards arrow on top close vertical bar space equals space open vertical bar straight lambda close vertical bar space space open vertical bar straight B with rightwards arrow on top space cross times space straight C with rightwards arrow on top close vertical bar space space space space space space space rightwards double arrow space space space open vertical bar straight A with rightwards arrow on top close vertical bar space equals space open vertical bar straight lambda close vertical bar space space open vertical bar straight B with rightwards arrow on top close vertical bar space open vertical bar straight C with rightwards arrow on top close vertical bar space space sin space straight pi over 6 space open vertical bar straight n with hat on top close vertical bar
where straight n with hat on top is perpendicular to straight B with rightwards arrow on top space as space well space as space straight C with rightwards arrow on top.
rightwards double arrow space space space space space 1 space equals space open vertical bar straight lambda close vertical bar space left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis space space open parentheses 1 half close parentheses space left parenthesis 1 right parenthesis space space space space space space space space space space space space space space space open square brackets because space space space straight A with rightwards arrow on top comma space straight B with rightwards arrow on top comma space straight C with rightwards arrow on top space are space unit space vectors close square brackets.
therefore space space space space space from space left parenthesis 1 right parenthesis comma space space space we space get comma space space straight A with rightwards arrow on top space space equals plus-or-minus 2 space open parentheses straight B with rightwards arrow on top space cross times space straight C with rightwards arrow on top close parentheses

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