Answer:
The given vector can be represented in unit vector as
[tex]\overrightarrow{w}=-30\widehat{i}+40\widehat{j}[/tex]
The magnitude of any vector [tex]\overrightarrow{r}=u\widehat{i}+v\widehat{j}[/tex] is given by
[tex]|w|=\sqrt{u^{2}+v^{2}}[/tex]
Applying values we get
[tex]|w|=\sqrt{-30^{2}+40^{2}}\\\\|r|=50[/tex]
We know that positive x axis in vertorial form is represented as
[tex]\overrightarrow{r}=\widehat{i}[/tex]
taking dot product of both the vector's we get
[tex]\overrightarrow{r}.\overrightarrow{w}=|r||w|cos(\theta )\\\\\therefore cos(\theta )=\frac{(-30\widehat{i}+40\widehat{j}).\widehat{i}}{50}\\\\\therefore \theta =cos^{-1}(\frac{-30}{50})=126.86^{o}[/tex]
Answer: [tex]||S||=50,0m[/tex], [tex]\theta=233,13\textdegree[/tex]
Explanation:
Let [tex]S=(-30.0 m, +40,0 m)[/tex], the magnitude is given by the following formula:
[tex]||S|| = \sqrt{(-30,0m)^{2}+(40,0m)^{2}}[/tex]
[tex]||S||=50,0m[/tex]
The angle between the direction of [tex]S[/tex] and the positive direction of the x axis is:
[tex]\theta = 180\textdegree- \tan^{-1} {\frac{40,0 m}{-30,0 m} }[/tex]
[tex]\theta=233,13\textdegree[/tex]
