Question

Let A = iA cos 0 + A sino be any vector. Another vector B which is normal to A is ?
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Answer 1

Answer:

Ā= îAcos€ + jAsin€

B is a vector normal to A

let,

B= xî + yj

here,

Ā.B

$= |a| |b| \cos\theta \\ = |a| |b| \cos90 \\ = |a| |b| \times 0 \\ = 0.....(1)$

Again,

A.B

$= a \cos \theta \times x + asin \theta \times y \\ = a(xcos \theta + ysin \theta).......(2)$

From (1) and (2), we get,

$a(xcos \theta + ysin \theta) = 0 \\ = > xcos \theta + ysin \theta = 0 \\ = > xcos \theta = - ysin \theta \\ = > \frac{x}{y} = - \frac{sin \theta}{cos \theta} \\ = > \frac{x}{y} = - \frac{bsin \theta}{ bcos \theta}$

therefore,

$x = bsin \theta \\ y = - bcos \theta$

$\therefore \: b = i \: bsin \theta - j \: bcos \theta$

ANS: C