Question

A vector has the components Ax=−31m and Ay=44m.

Answer 1

Magnitude of vector A is  51.4 m and the angle of vector A with positive x-axis θ is 127.1° .

The quantities which have direction as well as magnitude is known as vector quantities .

We are given that ,

Ax is vector component on x -axis=−31m

Ay is vector component on y -axis=44m

Therefore, the magnitude of the vectors $A_{x}$ and $A_{y}$ is $A$ which is given as,

$A = \sqrt{A_{x} ^{2} }+ {A_{y} ^{2} }\\A = \sqrt{{-31} ^{2} }+ {41 ^{2} }}\\A = 51.4m$

Thus, the magnitude of vector A will be 51.4 m

Now calculating the angle(θ) of the vector which make on the positive x axis, is given by the formula,

θ' = $tan^{-1}\frac{Ax}{Ay}$

θ' = $tan^{-1}\frac{-31}{44}$

θ' = $-59.9$°

Thus, here vector A is lies in 2nd quadrant which is  x-component of vector A that is negative and y-component is positive. So that from this to get the angle of vector with positive x-axis,

θ = 180° - θ'

θ=180 °- 52.9°

θ= 127.1°

Hence from above solution, the angle of vector A with positive x-axis θ is 127.1°

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