Respuesta :
The magnitude of the resulting vector, u – v, is approximately √34, and its angle of direction is approximately 78.1°.
What is a vector?
A vector is the type of quantity that has both magnitude and direction.
How to subtract two vectors?
Let the two vectors be [tex]\overrightarrow{u}= < u_{1},u_{2} >[/tex] and [tex]\overrightarrow{v}= < v_{1},v_{2} >[/tex]. We can subtract the two vectors as follows:
[tex]\overrightarrow{u}-\overrightarrow{v}= < u_{1}-v_{1},u_{2}-v_{2} >[/tex]
How to find the magnitude of a vector?
Let the vector be [tex]\overrightarrow{u}= < u_{1},u_{2} >[/tex]. The magnitude is given by:
[tex]\left| \overrightarrow{u} \right|=\sqrt{({u_{1})^{2}}+(u_{2})^{2}}[/tex]
How to find the angle between two vectors?
The angle between two vectors can be found using:
[tex]\cos\theta=\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{\left| \overrightarrow{u} \right|\cdot \left| \overrightarrow{v} \right|}[/tex]
- The two vectors are [tex]\overrightarrow{u}= < 5,2 >[/tex] and [tex]\overrightarrow{v}= < 2,-3 >[/tex]. By subtracting them, we get:
[tex]\overrightarrow{u}-\overrightarrow{v}= < 5,2 > - < 2,-3 > \\ = < 5-2,2-(-3) > \\= < 3,5 >[/tex]
- The magnitude of the resultant vector can be given by:
[tex]\left| < 3,5 > \right|=\sqrt{3^{2}+5^{2}}\\=\sqrt{9+25}\\=\sqrt{34}[/tex]
- To find the angle between the two vectors, we have to find the value of [tex]\overrightarrow{u}\cdot \overrightarrow{v}[/tex], [tex]\left| \overrightarrow{u}\right|[/tex], and [tex]\left| \overrightarrow{v}\right|[/tex].
[tex]\overrightarrow{u}\cdot \overrightarrow{v}= < 5,2 > \cdot < 2,-3 > \\=5(2)+2(-3)\\=10+(-6)\\=4[/tex]
[tex]\left| \overrightarrow{u} \right|=\sqrt{5^{2}+2^{2}}\\=\sqrt{25+4}\\=\sqrt{29}[/tex]
[tex]\left| \overrightarrow{v} \right|=\sqrt{2^{2}+(-3)^{2}}\\=\sqrt{4+9}\\=\sqrt{13}[/tex]
- The angle between the two vectors can be given by:
[tex]\cos\theta=\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{\left| \overrightarrow{u} \right|\cdot \left| \overrightarrow{v} \right|}\\=\frac{4}{\sqrt{29}\cdot \sqrt{13}}\\=\frac{4}{\sqrt{377}}=0.2060\\\theta=\cos^{-1}\left( 0.2060 \right)\\=78.1^\circ[/tex]
Therefore, the magnitude of the resulting vector, u – v, is approximately √34, and its angle of direction is approximately 78.1°.
Learn more about vectors here-https://brainly.com/question/11495559
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