Question

How do you find the magnitude and direction for $ U $ : magnitude $ 140 $ , bearing $ {160^o} $ $ V $ : magnitude $ 200 $ , bearing $ {290^o} $ ?

Answer 1

Hint: We are given two vectors in terms of magnitude and bearing. We have to simplify the given information to find the magnitude and direction for the vectors. The direction of a vector is represented in terms of angle from a reference axis. Using the bearing of the given vector we can find the direction of the vector.

Complete step by step solution:
We have been given two vectors,
 $ U $ : magnitude $ 140 $ , bearing $ {160^o} $
 $ V $ : magnitude $ 200 $ , bearing $ {290^o} $
The magnitude of the vectors are already given. Using the bearing of the given vectors we can find the direction of the vectors.
The bearing is given in terms of angles which the vector makes anti-clockwise with the x-axis.
We can represent a vector in terms of magnitude and bearing as,
We can write the vector $ U $ as $ U = 140\left( {\cos {{160}^o},\sin {{160}^o}} \right) $
The magnitude of vector $ U $ is $ 140\;units $ and the direction of the vector $ U $ is that the vector makes $ {160^o} $ anti-clockwise from the positive x-axis or we can say $ {\left( {180 - 160} \right)^o} = {20^o} $ clockwise from the negative x-axis.
Similarly, we can write the vector $ V $ as $ V = 200\left( {\cos {{290}^o},\sin {{290}^o}} \right) $
The magnitude of vector $ V $ is $ 200\;units $ and the direction of the vector $ U $ is that the vector makes $ {290^o} $ anti-clockwise from the positive x-axis or we can say $ {\left( {360 - 290} \right)^o} = {70^o} $ clockwise from the positive x-axis.

Note: A quantity having both magnitude and direction is known as vector quantity. In the given question we were already given the magnitude of the vectors. We found the direction of the vectors using the given bearings which means the angle which the vector makes with the positive x-axis anti-clockwise. The sense of anti-clockwise and clockwise is important to know the direction of the vector correctly.