Question

vector

a⃗

of length 8 m lies at an angle of

60o

above the x-axis in the first quadrant. vector

b⃗

of length 5 m lies

53o

below the x-axis in the fourth quadrant. determine the magnitude of

a⃗ −b⃗

9.94 m

7.60 m

10.96 m

20.21 m

Answer 1

Answer on Question #56255-Physics-Mechanics-Relativity

Vector $\vec{a}$ of length 8 m lies at an angle of $60{}^{\circ}$ above the x-axis in the first quadrant. Vector $\vec{b}$ of length 5 m lies $53{}^{\circ}$ below the x-axis in the fourth quadrant. Determine the magnitude of $\vec{a} - \vec{b}$

9.94 m

7.60 m

10.96 m

20.21 m

Solution

$$\vec{a} = (a \cos 60{}^{\circ}; a \sin 60{}^{\circ}) = (8 \cos 60{}^{\circ}; 8 \sin 60{}^{\circ}) = (4; 6.93).$$ $$\vec{b} = (b \cos 53{}^{\circ}; -b \sin 53{}^{\circ}) = (5 \cos 53{}^{\circ}; -5 \sin 53{}^{\circ}) = (3.01; -3.99).$$ $$\vec{a} - \vec{b} = (4 - 3.01; 6.93 - (-3.99)) = (0.99; 10.92).$$

The magnitude of $\vec{a} - \vec{b}$ is

$$|\vec{a} - \vec{b}| = \sqrt{0.99^2 + 10.92^2} = 10.96 \, \text{m}$$

Answer: 10.96 m.

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