if vector A=i cap *A cos thita + j cap *A sin thita, then another vector B , WHICH IS PERPENDICULAR TO VECTOR A,can be expressed as?give detailed answer?
To find a vector that is perpendicular to vector A, we can use the concept of the dot product. The dot product of two vectors is zero if and only if they are perpendicular to each other.
Let's start by representing vector A in terms of its components. The expression A = i cap * A cos(theta) + j cap * A sin(theta) can be written as:
A = (A cos(theta)) * i cap + (A sin(theta)) * j cap
Now, let's find vector B using the dot product with this formula:
B · A = 0
Since vector B is perpendicular to vector A, the dot product of B and A will be zero.
B · [(A cos(theta)) * i cap + (A sin(theta)) * j cap] = 0
Expanding this expression:
(Bx * i cap + By * j cap) · [(A cos(theta)) * i cap + (A sin(theta)) * j cap] = 0
Now we can equate the components to zero individually:
(Bx * i cap) · [(A cos(theta)) * i cap + (A sin(theta)) * j cap] + (By * j cap) · [(A cos(theta)) * i cap + (A sin(theta)) * j cap] = 0
(Bx * (A cos(theta))) + (By * (A sin(theta))) = 0
We have two equations here:
Bx * (A cos(theta)) + By * (A sin(theta)) = 0 ..........(1)
and Bx * (A sin(theta)) - By * (A cos(theta)) = 0 ..........(2)
To find vector B, we solve these equations simultaneously.
Divide equation (1) by (A cos(theta)):
Bx = - (By * (A sin(theta))) / (A cos(theta))
This gives us the value of Bx. Now we substitute this value into equation (2):
-(By * (A sin(theta))) / (A cos(theta))) * (A sin(theta)) - By * (A cos(theta)) = 0
Simplifying this equation:
- By * (A sin^2(theta) / (A cos(theta)) - By * A cos(theta) = 0
By * [ -A sin^2(theta) - A cos^2(theta)] = 0
By * [-A (sin^2(theta) + cos^2(theta))] = 0
Since sin^2(theta) + cos^2(theta) = 1, we can simplify further:
By * (-A) = 0
By = 0
Therefore, By = 0. This means that the y-component of vector B is zero. Now we substitute this value back into equation (1) to find Bx:
Bx = - (0 * (A sin(theta))) / (A cos(theta)) = 0
Therefore, Bx = 0.
In conclusion, vector B can be expressed as B = i cap * 0 + j cap * 0, which simplifies to B = 0.
A = |A| cos thita i + |A| sin thita j
B = x i + y j
A dot B = 0 for perpendicular
|A| x cos thita + |A| y sin thita = 0
if x = -sin thita and y = cos thita
that works
(also the opposite of course)