Let $\to A =^i A cos θ +^j A sin θ$ be any vector. Another vector $\to B$ , which is normal to $A$ can be expressed as

^i B cos θ -^j B sin θ

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^i B cos θ +^j B sin θ

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^i B sin θ -^j B cos θ

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^i B sin θ +^j B cos θ

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Solution

The correct option is C $^i B sin θ -^j B cos θ$
$\to A =^i A c o s θ +^j A s i n θ$ and consider $\to A =^i x +^j y$ . Now, B is normal to A then:
$\to A . \to B = 0$
$\to A . \to B$ $= (^i A c o s θ +^j A s i n θ) . (^i x +^j y)$
$= A x c o s θ + A y s i n θ$
Therefore, for $\to A . \to B = 0$ x should be of form $B s i n θ$ and y of form $- B c o s θ$ .

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\to A =^i A c o s θ +^j A s i n θ

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\to A =^i A c o s θ +^j A s i n θ

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Let $¯ A = A cos θ^i + A sin θ^j$ be any vector. Another vector $¯ B$ , which is normal to $¯ A$ can be expressed as

\to A = A cos θ^i + A sin θ^j

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\to a, \to b, \to c

[\to a \to b \to c] \neq

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\to r

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