If vectors A= cos Omega t i + sin Omega t j and B = cos Omega t/2i + sin Omega t/2 j are function of time then the value of the at which they are perpendicular to each other

To determine when the vectors A and B are perpendicular, we need to understand the concept of the dot product. Two vectors are perpendicular when their dot product equals zero. Let's break down the problem step by step.

Defining the Vectors

First, let’s write down the vectors clearly:

• A = cos(Ωt) i + sin(Ωt) j
• B = cos(Ωt/2) i + sin(Ωt/2) j

Calculating the Dot Product

The dot product of two vectors A and B is given by the formula:

A · B = A_x * B_x + A_y * B_y

Substituting the components from vectors A and B:

• A_x = cos(Ωt)
• A_y = sin(Ωt)
• B_x = cos(Ωt/2)
• B_y = sin(Ωt/2)

Thus, the dot product can be expressed as:

A · B = cos(Ωt) * cos(Ωt/2) + sin(Ωt) * sin(Ωt/2)

Using Trigonometric Identities

We can apply the cosine addition formula, which states that:

cos(a - b) = cos(a)cos(b) + sin(a)sin(b)

In our case, we can rewrite the dot product as:

A · B = cos(Ωt - Ωt/2) = cos(Ωt/2)

Setting the Dot Product to Zero

For vectors A and B to be perpendicular, we need the dot product to equal zero:

cos(Ωt/2) = 0

The cosine function equals zero at odd multiples of π/2. Therefore, we can write:

Ωt/2 = (2n + 1)π/2, where n is any integer.

Finding the Time 't'

To isolate t, we can rearrange the equation:

t = (2(2n + 1)π) / (Ω)

This shows that the vectors A and B will be perpendicular at specific times determined by the integer n.

Examples of Specific Values

For instance, if we take n = 0, we find:

t = (2π) / Ω

If n = 1, then:

t = (6π) / Ω

These values demonstrate just a few instances of when the vectors A and B are perpendicular during their motion over time.

In summary, whenever you want to check if two time-dependent vectors are perpendicular, the key step is calculating their dot product and setting it to zero. By doing so, you can derive the specific instances where this condition holds true.