Answer :
Answer:
(a)
[tex]|\vec{A} + \vec{B}| = 31.3~{\rm units}\\\theta = 333.4^\circ[/tex]
(b)
[tex]|\vec{A} + \vec{B}| = 31.3~{\rm units}[/tex]
(c)
[tex]|\vec{A} - \vec{B}| = 31.3~{\rm units}\\\theta =206.56^\circ[/tex]
(d)
[tex]|\vec{A} - \vec{B}| = 31.3~{\rm units}[/tex]
(e)
[tex]|\vec{B} - \vec{A}| = 31.3~{\rm units}\\\theta =26.5^\circ[/tex]
Explanation:
[tex]\vec{A} = -14\^y\\\vec{B} = 28\^x[/tex]
(a)
[tex]\vec{A} + \vec{B} = 28\^x - 14\^y\\|\vec{A} + \vec{B}| = \sqrt{28^2 + (-14^2)} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = -0.5\\\theta = \arctan(-0.5) = 333.4^\circ[/tex]
Here, the vector is in the fourth quadrant. Hence, we subtracted the arctan value from 360°, because the arctan value gives the angle from the x-axis.
(b)
[tex]|\vec{A} + \vec{B}| = \sqrt{28^2 + (-14^2)} = 31.3~{\rm units}[/tex]
(c)
[tex]\vec{A} - \vec{B} = -28\^x - 14\^y\\|\vec{A} - \vec{B}| = \sqrt{(-28)^2 + (-14^2)} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = 0.5\\\theta = \arctan(0.5) = 206.56^\circ[/tex]
Here, the vector is in the third quadrant. Hence, we added 180° to the arctan value.
(d)
[tex]|\vec{A} - \vec{B}| = \sqrt{(-28)^2 + (-14^2)} = 31.3~{\rm units}[/tex]
(e)
[tex]\vec{B} - \vec{A} = 28\^x + 14\^y\\|\vec{B} - \vec{A}| = \sqrt{28^2 + 14^2} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = 0.5\\\theta = \arctan(0.5) = 26.5^\circ[/tex]
Here, the vector is in the first quadrant.