Question

Vector A points in the negative y-direction and has a magnitude of 14 units. Vector B has twice the magnitude and points in the positive x-direction.
(a) Find the direction and magnitude of A + B. (degrees counterclockwise from the +x axis)
(b) |A + B| = ?
(c) Find the direction and magnitude of A - B. (degrees counterclockwise from the +x axis)
(d) |A - B| =?
(e) Find the direction and magnitude of B - A. (degrees counterclockwise from the +x axis)

Answer 1

Answer:

(a)

$|\vec{A} + \vec{B}| = 31.3~{\rm units}\\\theta = 333.4^\circ$

(b)

$|\vec{A} + \vec{B}| = 31.3~{\rm units}$

(c)

$|\vec{A} - \vec{B}| = 31.3~{\rm units}\\\theta =206.56^\circ$

(d)

$|\vec{A} - \vec{B}| = 31.3~{\rm units}$

(e)

$|\vec{B} - \vec{A}| = 31.3~{\rm units}\\\theta =26.5^\circ$

Explanation:

$\vec{A} = -14\^y\\\vec{B} = 28\^x$

(a)

$\vec{A} + \vec{B} = 28\^x - 14\^y\\|\vec{A} + \vec{B}| = \sqrt{28^2 + (-14^2)} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = -0.5\\\theta = \arctan(-0.5) = 333.4^\circ$

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)

$|\vec{A} + \vec{B}| = \sqrt{28^2 + (-14^2)} = 31.3~{\rm units}$

(c)

$\vec{A} - \vec{B} = -28\^x - 14\^y\\|\vec{A} - \vec{B}| = \sqrt{(-28)^2 + (-14^2)} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = 0.5\\\theta = \arctan(0.5) = 206.56^\circ$

Here, the vector is in the third quadrant. Hence, we added 180° to the arctan value.

(d)

$|\vec{A} - \vec{B}| = \sqrt{(-28)^2 + (-14^2)} = 31.3~{\rm units}$

(e)

$\vec{B} - \vec{A} = 28\^x + 14\^y\\|\vec{B} - \vec{A}| = \sqrt{28^2 + 14^2} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = 0.5\\\theta = \arctan(0.5) = 26.5^\circ$

Here, the vector is in the first quadrant.