If vectors and are functions of time, then the value of at which they are orthogonal to each other is [2015]

Question

If vectors $\vec{A} = \cos ω t \hat{i} + \sin ω t \hat{j}$ and $\vec{B} = \cos \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j}$ are functions of time, then the value of $t$ at which they are orthogonal to each other is [2015]

Smart Achievers · Accepted Answer

(1)

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 $\vec{B} = \cos \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j}$

$∴$ $\vec{A} \cdot \vec{B} = (\cos ω t \hat{i} + \sin ω t \hat{j}) \cdot (\cos \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j})$

$= \cos ω t \cos \frac{ω t}{2} + \sin ω t \sin \frac{ω t}{2} = \cos (ω t - \frac{ω t}{2})$

But $\vec{A} \cdot \vec{B} = 0$ (as $\vec{A}$ and $\vec{B}$ are orthogonal to each other)

$∴$ $\cos (ω t - \frac{ω t}{2}) = 0$

$\cos (ω t - \frac{ω t}{2}) = \cos \frac{π}{2} or ω t - \frac{ω t}{2} = \frac{π}{2}$

$\frac{ω t}{2} = \frac{π}{2} or t = \frac{π}{ω}$

Solution

Q.
If vectors $\vec{A} = \cos ω t \hat{i} + \sin ω t \hat{j}$ and $\vec{B} = \cos \frac{ω t}{2} \hat{i} + \sin \frac{ω t}{2} \hat{j}$ are functions of time, then the value of $t$ at which they are orthogonal to each other is [2015]

1 $t = \frac{π}{ω}$

2 $t = 0$

3 $t = \frac{π}{4 ω}$

4 $t = \frac{π}{2 ω}$

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Solution

Q. If vectors A→=cosωt i^+sinωt j^ and B→=cosωt2i^+sinωt2j^ are functions of time, then the value of t at which they are orthogonal to each other is [2015]

1 t=πω

2 t=0

3 t=π4ω