Vector which is perpendicular to (acosθ)i^+(bsinθ)j^ is given by :

A
$(b \sin θ) \hat{i} - (a \cos θ) \hat{j}$
B
$(\frac{1}{a} \sin θ) \hat{i} - (\frac{1}{b} \cos θ) \hat{j}$
C
$5 \hat{k}$
D
All of these

Solution:

$(i) [(a \cos θ) \hat{i} + (b \sin θ) \hat{j}] . [(b \sin θ) \hat{i} - (a \cos θ) \hat{j}]$

$= a b \sin θ \cos θ - b a \sin θ \cos θ = 0$ $(i i) [a (\cos θ) \hat{i} + (b \sin θ) \hat{j}] . [(\frac{1}{a} \sin θ) \hat{i} - (\frac{1}{b} \cos θ) \hat{j}]$

$= \sin θ \cos θ - \sin θ \cos θ = 0$

$(i i i) [(a \cos θ) \hat{i} + (b \sin θ) \hat{j}] .5 \hat{k} = 0$

Hence, all the three options are correct because the dot product of two perpendicular Also Check: is zero

Vector which is perpendicular to $(a \cos θ) \hat{i} + (b \sin θ) \hat{j}$ is given by :

Solution:

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