First slide

Vectors

Question

If vector $\vec{A} = cosωt \hat{i} + sinωt \hat{j}$ and $\vec{B} = \cos \frac{ωt}{2} \hat{i} + \sin \frac{ωt}{2} \hat{j}$ are functions of time, then the value of t at which they are orthogonal to each other is

Difficult

A

$t = \frac{π}{ω}$
B

$t = 0$
C

$t = \frac{π}{4 ω}$
D

$t = \frac{π}{2 ω}$

Solution

$Two vectors \vec{A} and \vec{B} are orthogonal to each other, if$ $their scalar product is zero i.e. \vec{A} \cdot \vec{B} = 0 .$
$Here, \vec{A} = \cos ω t \hat{i} + \sin ω t \hat{j}$ $and \vec{B} = \cos \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j}$
$∴ \vec{A} \cdot \vec{B} = (\cos ω t \hat{i} + \sin ω t \hat{j}) \cdot (\cos \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j})$
$= \cos ω t \cos \frac{ω t}{2} + \sin ω t \sin \frac{ω t}{2}$
$(∵ \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = 1 and \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{i} = 0)$
$= \cos (ω t - \frac{ω t}{2})$
$(∵ \cos (A - B) = \cos A \cos B + \sin A \sin B)$
$But \vec{A} \cdot \vec{B} = 0 (as \vec{A} and \vec{B} are orthogonal to each other)$
$∴ \cos (ω t - \frac{ω t}{2}) = 0$
$\cos (ω t - \frac{ω t}{2}) = \cos \frac{π}{2} or ω t - \frac{ω t}{2} = \frac{π}{2}$
$\frac{ω t}{2} = \frac{π}{2} or t = \frac{π}{ω}$

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