Vector A has a magnitude of 3.0 units and makes an angle of −90.0º with the positive x-axis. Vector B has a magnitude of 4.0 units and makes an angle of −120º with the positive x-axis. What is the magnitude of the vector sum A + B?
A) 1.0 units
B) 6.8 units
C) 7.0 units
D) 5.0 units

Question

Quizplus · Accepted Answer

To find the magnitude of the vector sum $A + B$, we first break down vectors $A$ and $B$ into their components. For vector $A$, since it makes an angle of $-90.0^\circ$ with the positive x-axis, it is directed entirely along the negative y-axis. Thus, its components are $A_x = 0$ and $A_y = -3.0$ units.For vector $B$, with a magnitude of 4.0 units and an angle of $-120^\circ$ from the positive x-axis, we calculate its components using trigonometry. The angle $-120^\circ$ is equivalent to $240^\circ$ in standard position, which places it in the third quadrant where both x and y components are negative. The components are:- $B_x = 4.0 \cos(-120^\circ) = 4.0 \cos(240^\circ) = -2$ units (since $\cos(240^\circ) = -1/2$)- $B_y = 4.0 \sin(-120^\circ) = 4.0 \sin(240^\circ) = -2\sqrt{3}$ units (since $\sin(240^\circ) = -\sqrt{3}/2$)The sum of the vectors $A + B$ in component form is:- $x$-component: $A_x + B_x = 0 - 2 = -2$ units- $y$-component: $A_y + B_y = -3.0 - 2\sqrt{3}$ unitsThe magnitude of the vector sum is calculated using the Pythagorean theorem:$$|A + B| = \sqrt{(-2)^2 + (-3.0 - 2\sqrt{3})^2} = \sqrt{4 + 9 + 12\sqrt{3} + 12} = \sqrt{25 + 12\sqrt{3}}$$Approximating $\sqrt{3} \approx 1.732$, we get:$$|A + B| \approx \sqrt{25 + 12(1.732)} = \sqrt{25 + 20.784} = \sqrt{45.784} \approx 6.8$$ units.

Vector a Has a Magnitude of 3