The magnitude and the direction of the resulting vector is 3 and 0°, respectively.
Procedure - Determination of a vector by subtraction
Let assume that angles are in standard position. According to the statement, we have following model for the direction of the resulting vector in terms of the vectors known in statement:
[tex]\vec R = \vec K - \vec H[/tex] (1)
Where [tex]\vec R[/tex] is the resulting vector.
If we know that [tex]\vec H = 3\cdot (\cos 90^{\circ}, \sin 90^{\circ})[/tex], [tex]\vec K = k\cdot (\cos 45^{\circ}, \cos 45^{\circ})[/tex] and [tex]\vec R = r\cdot (\cos 0^{\circ}, \sin 0^{\circ})[/tex], then we have the following expression:
[tex]r\cdot (1, 0) = k \cdot \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)-3\cdot (0, 1)[/tex]
[tex](r, 0) = \left(\frac{\sqrt{2}\cdot k}{2}, \frac{\sqrt{2}\cdot k}{2} - 3 \right)[/tex]
Then we have the following system of linear equations:
[tex]r = \frac{\sqrt{2}\cdot k}{2}[/tex] (2)
[tex]0 = \frac{\sqrt{2}\cdot k}{2}-3[/tex] (3)
By (3):
[tex]k = 3\sqrt{2}[/tex]
By (2):
[tex]r = 3[/tex]
Hence, the magnitude and the direction of the resulting vector is 3 and 0°, respectively. [tex]\blacksquare[/tex]
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