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Vector Algebra

Question
CBSEENMA12034001
Wired Faculty App

Prove that if straight u with rightwards arrow on top space equals space straight u subscript 1 space straight i with hat on top space plus space straight u subscript 2 space straight j with hat on top space and space straight v with rightwards arrow on top space equals space straight v subscript 1 space straight i with hat on top space plus space straight v subscript 2 space straight j with hat on top are non-zero vectors, then they are parallel if and only if u2v2 – u2v1 = 0.

Solution

Let straight u with rightwards arrow on top space equals space straight u subscript 1 straight i with hat on top space plus space straight u subscript 2 straight j with hat on top comma space space straight v with rightwards arrow on top space equals space straight v subscript 1 straight i with hat on top space plus space straight v subscript 2 straight j with hat on top be parallel vectors. 
therefore space space space there space exists space straight a space non minus zero space scalar space straight k space such space that space straight u with rightwards arrow on top space equals space straight k space straight v with rightwards arrow on top therefore space space space space straight u subscript 1 straight i with hat on top space plus space straight u subscript 2 straight j with hat on top space equals space straight k left parenthesis straight v straight i with hat on top space plus space straight v subscript 2 straight j with hat on top right parenthesis or space space left parenthesis straight u subscript 1 space space minus space kv subscript 1 right parenthesis space equals space 0 comma space space space straight u subscript 2 space minus kv subscript 2 space equals space 0 therefore space space space space straight k space equals space straight u subscript 1 over straight v subscript 1 comma space space space straight k space equals space straight u subscript 2 over straight v subscript 2 Equating space values space of space straight k comma space we space get comma space space space space space space space space space space space space space space space space space space space space straight u subscript 1 over straight v subscript 1 space equals space straight u subscript 2 over straight v subscript 2 therefore space space space space space space space space straight u subscript 1 straight v subscript 2 space equals space straight u subscript 2 straight v subscript 1 space space space space space space space space space rightwards double arrow space space space space space straight u subscript 1 straight v subscript 2 space minus space straight u subscript 2 space straight v subscript 1 space equals space 0
Similarly, if straight u subscript 1 space straight v subscript 2 space equals space straight u subscript 2 space straight v subscript 1 space space space space rightwards double arrow space space space space space straight u subscript 1 space straight v subscript 2 space minus space straight u subscript 2 space straight v subscript 1 space equals space 0
i.e.,       straight u with rightwards arrow on top comma space space straight v with rightwards arrow on top are parallel
Hence the result.
   

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