Question

A vector has the components Ax=−31m and Ay=41m. What is the magnitude of this vector? What angle does this vector make with the positive x axis?

Answer 1

Answer:

Magnitude of vector A is  51.4 m

The angle of vector A with positive x-axis is θ= $127.1^o}$

Explanation:

Given is  

$A_{x}$=-31m

$A_{y$=41m

Let $A_{x}$ and $A_{y$ are components of a vector A,

The magnitude of vector A will be:

$A=\sqrt{A_{x}^2 +A_{y}^2}$

   =  $\sqrt{(-31)^{2} +41^2}$

A  = 51.4 m

we know

 θ1= $tan^{-1} \frac{A_{y} }{A_{x} }$

   = $tan^{-1}(-\frac{41}{31} )$

   =$-52.9^{o$

Vector A is present in 2nd quadrant as x-component of vector A is negative and y-component is positive.

For finding the angle of vector with positive x-axis,

θ=180-52.9

θ= $127.1^o}$