Question

Vector A S is 3.00 units in length and points along the positive x - axis. Vector B S is 4.00 units in length and points along the negative y - axis. Use graphical methods to find the magnitude and direction of the vectors (a) A S 1 B S and (b) A S 2 B S .

Answer 1

Answers:

a) Magnitude:5 units  Direction:$-53.13\°$ or $53.13\°$ South of East

b) Magnitude:5 units  Direction:$53.13\°$ or $53.13\°$ North of East

Explanation:

Vectors in two dimensions have two components Y and X, and are written in the following form:

(X,Y)

In this case we have two vectors:

$\vec{A}=(3,0)$

$\vec{B}= (0,-4)$

We need to find $\vec{A} + \vec{B}$ and $\vec{A} - \vec{B}$, their magnitude and direction.

For the magnitude we will use the formula to calculate the distance $d$ between two points (pithagorean theorem):

$d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}$ (1)

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:  

$tan \theta=\frac{Y2-Y1}{X2-X1}$ (2)

Now, let's begin with the answers:

a) $\vec{A} + \vec{B}$

Adding both vectors:

$\vec{A} + \vec{B}=(3,0) + (0,-4)=(3,-4)$

Finding the magnitude:

$d_{a}=\sqrt{{(-4-0)}^{2} +{(0-3)}^{2}}$

$d_{a}=\sqrt{{16+9}=\sqrt{{125}$

$d_{a}=5 units$ This is the magnitude of  $\vec{A} + \vec{B}$

Finding the direction:

$tan \theta=\frac{-4}{3}$

$\theta=tan^{-1} (\frac{-4}{3})$

$\theta=-53.13\°$ This is the direction of  $\vec{A} + \vec{B}$

If we assume the positive x axis as the East and the negative y axis as the South, we can also express this direction as:

$53.13\°$ South of East

b) $\vec{A} - \vec{B}$

Substracting $\vec{B}$ to $\vec{A}$:

$\vec{A} - \vec{B}=(3,0) - (0,-4)=(3,4)$

Finding the magnitude:

$d_{a}=\sqrt{{(-4-0)}^{2} +{(0-3)}^{2}}$

$d_{a}=\sqrt{{16+9}=\sqrt{{125}$

$d_{a}=5 units$ This is the magnitude of  $\vec{A} - \vec{B}$

Finding the direction:

$tan \alpha=\frac{4}{3}$

$\alpha=tan^{-1} (\frac{4}{3})$

$\alpha=53.13\°$ This is the direction of  $\vec{A} - \vec{B}$

If we assume the positive x axis as the East and the positivr y axis as the North, we can also express this direction as:

$53.13\°$ North of East