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Algebra of Complex Numbers

Let A̅ =A c...

Question

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

^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 D $^i B sin θ -^j B cos θ$
$\to A . \to B$ must be zero, which is the case for option C only. Dot products of different options are given as:
A. $A B {c o s}^{2} θ - A B {s i n}^{2} θ$ , not zero.
B. $A B {c o s}^{2} θ + A B {s i n}^{2} θ = A B$ , not zero.
C. $A B c o s θ s i n θ - A B s i n θ c o s θ = 0$
D. $A B c o s θ s i n θ + A B s i n θ c o s θ$ , not zero

Suggest Corrections

Similar questions

\to A =^i A c o s θ +^j A s i n θ

\to B

\to A

\to A = A cos θ^i + A sin θ^j

\to B

\to A

Let $\to A = A c o s θ^i + A s i n θ^j$ be any vector.Then another vector $\to B$ which is normal to $\to A$ can be?

¯ a, ¯ b, ¯ c

¯ a

¯ b + ¯ c

¯ b

¯ c + ¯ a

¯ c

¯ a + ¯ b

∣ ∣ ¯ a + ¯ b + ¯ c ∣ ∣

\to a

\to b

\frac{{∣ ∣ ¯ a + ¯ b ∣ ∣}^{2} - {∣ ∣ ¯ a - ¯ b ∣ ∣}^{2}}{{∣ ∣ ¯ a + ¯ b ∣ ∣}^{2} + {∣ ∣ ¯ a - ¯ b ∣ ∣}^{2}}

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