Question

Find the direction and magnitude of the vectors (a) A = (25m)x + (-12m)y(b) B = (2.0m)x + (15m)y and(c) A + B

Answer 1

ANSWER

$\begin{gathered} a)\theta=-25.6\degree;|A|=27.73m \\ b)\theta=82.4\degree;|B|=15.13m \\ c)\theta=6.3\degree^{};|A+B|=27.17m \end{gathered}$

EXPLANATION

To find the direction of a two-dimensional vector, we apply the formula:

$\theta=\tan ^{-1}(\frac{a_y}{a_x})$

To find the magnitude of a two-dimensional vector, we apply the formula:

$|a|=\sqrt[]{(a^{}_y)^2+(a_x)^2}$

a) The direction of vector A is:

$\begin{gathered} \theta=\tan ^{-1}(\frac{-12m}{25m}) \\ \theta=\tan ^{-1}(-0.48) \\ \theta=-25.6\degree \end{gathered}$

The magnitude of vector A is:

$\begin{gathered} |A|=\sqrt[]{(25m)^2+(-12m)^2} \\ |A|=\sqrt[]{625m^2+144m^2}=\sqrt[]{769m^2} \\ |A|=27.73m \end{gathered}$

b) The direction of vector B is:

$\begin{gathered} \theta=\tan ^{-1}(\frac{15m}{2m}) \\ \theta=\tan ^{-1}(7.5) \\ \theta=82.4\degree \end{gathered}$

The magnitude of vector B is:

$\begin{gathered} |B|=\sqrt[]{(2.0m)^2+(15m)^2} \\ |B|=\sqrt[]{4m^2+225m^2}=\sqrt[]{229m^2} \\ |B|=15.13m \end{gathered}$

c) First, we have to find the sum of A and B:

$\begin{gathered} A+B=(25m)x+(-12m)y+(2.0m)x+(15m)y \\ A+B=(25m+2.0m)x+(-12m+15m)y \\ A+B=(27m)x+(3m)y \end{gathered}$

The direction of the vector (A + B) is:

$\begin{gathered} \theta=\tan ^{-1}(\frac{3m}{27m}) \\ \theta=\tan ^{-1}(0.1111) \\ \theta=6.3\degree^{} \end{gathered}$

The magnitude of the vector (A + B) is:

$\begin{gathered} |A+B|=\sqrt[]{(27m)^2+(3m)^2} \\ |A+B|=\sqrt[]{729m^2+9m^2}=\sqrt[]{738m^2} \\ |A+B|=27.17m \end{gathered}$