Let overline A = hat i A cos θ+ hat j A sin θ, be any vector. Another vector B which is normal to A is:

Question

Let overline A = hat i A cos θ+ hat j A sin θ, be any vector. Another vector B which is normal to A is: (A)  hat i B cos θ+ hat j B sin θ (B)  hat i B sin θ+ hat j B cos θ (C)  hat i B sin θ- hat j B cos θ (D)  hat i A cos θ- hat j A sin θ

Tardigrade Admin · Accepted Answer

For perpendicular condition, A ⋅ B =0 A ⋅ B =( hat i A cos θ+ hat j A sin θ)( hat i B sin θ- hat j B cos θ) = A cos θ B sin θ- A sin θ B cos θ = AB sin θ cos θ- AB sin θ cos θ =0 A ⋅ B =0 (condition satisfied)

Q. Let $\overline{A} = \hat{i} A cos θ + \hat{j} A sin θ,$ be any vector. Another vector $B$ which is normal to $A$ is:

$\hat{i} B cos θ + \hat{j} B sin θ$

$\hat{i} B sin θ + \hat{j} B cos θ$

$\hat{i} B sin θ - \hat{j} B cos θ$

$\hat{i} A cos θ - \hat{j} A sin θ$

For perpendicular condition,
$A \cdot B = 0$
$A \cdot B$
$= (\hat{i} A cos θ + \hat{j} A sin θ) (\hat{i} B sin θ - \hat{j} B cos θ)$
$= A cos θB sin θ - A sin θB cos θ$
$= A B sin θ cos θ - A B sin θ cos θ$
$= 0$
$A \cdot B = 0$ (condition satisfied)

Q. Let A=i^Acosθ+j^​Asinθ, be any vector. Another vector B which is normal to A is:

i^Bcosθ+j^​Bsinθ

i^Bsinθ+j^​Bcosθ

i^Bsinθ−j^​Bcosθ

i^Acosθ−j^​Asinθ