Answer:
The magnitude of the resultant is approximately 5.391 units.
Explanation:
We notice that vectors have the following characteristics: A magnitude and direction. That is:
[tex]\vec v = r\cdot (\cos \theta\,\hat{i}+\sin \theta \,\hat{j})[/tex]
Where:
[tex]r[/tex] - Magnitude, dimensionless.
[tex]\theta[/tex] - Direction, measured in radians.
This expression can be rearranged as:
[tex]\vec v = r\cdot \cos \theta \,\hat{i}+r\cdot \sin \theta \,j[/tex]
[tex]\vec v = v_{x}\,\hat {i}+v_{y}\,\hat{j}[/tex]
Where:
[tex]v_{x} = r\cdot \cos \theta[/tex]
[tex]v_{y} = r\cdot \sin \theta[/tex]
If we know a set of vectors of the form [tex]\vec {v} = v_{x}\,\hat{i}+v_{y}\,\hat{j}[/tex], the resultant vector ([tex]\vec {R}[/tex]) is the vectorial sum of every vector in the set:
[tex]\vec{R} = \Sigma_{i=1}^{n} v_{i,x}\,\hat{i}+\Sigma_{i=1}^{n}v_{i,y} \,\hat{j}[/tex]
If we know that [tex]\vec{A} = 2.5\,\hat{i}+4.3\,\hat{j}[/tex], [tex]\vec {B} = 6.1\,\hat{i}-2.1\,\hat{j}[/tex], [tex]\vec C = -3.6\,\hat{i}+1.0\,\hat{j}[/tex] and [tex]\vec{D} = -1.5\,\hat{i}-7.3\,\hat{j}[/tex], the resultant vector is:
[tex]\vec {R} = \vec {A}+\vec {B} + \vec {C} + \vec {D}[/tex]
[tex]\vec {R}=(2.5+6.1-3.6-1.5)\,\hat{i} + (4.3-2.1+1.0-7.3)\,\hat{j}[/tex]
[tex]\vec {R} = 3.5\,\hat{i} -4.1\,\hat{j}[/tex]
And lastly, we obtain the magnitude of the resultant by Pythagoras' Theorem:
[tex]\|\vec R\| = \sqrt{3.5^{2}+(-4.1)^{2}}[/tex]
[tex]\|\vec {R}\| \approx 5.391[/tex]
The magnitude of the resultant is approximately 5.391 units.
