Question

question 4

1. given the vectors \\(\vec{a}=6\hat{i} - 4\hat{j}\\) and \\(\vec{b}=(10,60^{circ})\\)
2. determine \\(\vec{a}+\vec{b}\\) (10pts)
3. determine \\(\vec{a}-\vec{b}\\) (10pts)

*note that vector b is in polar form. express your final answers in polar form (i.e., find the magnitude and direction angle).

Explanation:

Step1: Add vectors component - wise

Given $\vec{A}=6\hat{i}-4\hat{j}$ and $\vec{B}=(10,60^{\circ})$. First convert $\vec{B}$ to rectangular - form. The $x$ - component of $\vec{B}$ is $B_x = 10\cos60^{\circ}=10\times\frac{1}{2}=5$ and the $y$ - component of $\vec{B}$ is $B_y = 10\sin60^{\circ}=10\times\frac{\sqrt{3}}{2}=5\sqrt{3}$. Then $\vec{B}=5\hat{i}+5\sqrt{3}\hat{j}$. $\vec{A}+\vec{B}=(6 + 5)\hat{i}+(-4 + 5\sqrt{3})\hat{j}=11\hat{i}+(5\sqrt{3}-4)\hat{j}$.

Step2: Find magnitude of $\vec{A}+\vec{B}$

The magnitude of a vector $\vec{R}=R_x\hat{i}+R_y\hat{j}$ is $|\vec{R}|=\sqrt{R_x^{2}+R_y^{2}}$. For $\vec{R}=\vec{A}+\vec{B}$, $R_x = 11$ and $R_y=5\sqrt{3}-4$. So $|\vec{A}+\vec{B}|=\sqrt{11^{2}+(5\sqrt{3}-4)^{2}}=\sqrt{121+(75 - 40\sqrt{3}+16)}=\sqrt{121 + 75+16-40\sqrt{3}}=\sqrt{212-40\sqrt{3}}\approx\sqrt{212 - 40\times1.732}=\sqrt{212 - 69.28}=\sqrt{142.72}\approx11.94$.

Step3: Find direction angle of $\vec{A}+\vec{B}$

The direction angle $\theta$ of a vector $\vec{R}=R_x\hat{i}+R_y\hat{j}$ is given by $\tan\theta=\frac{R_y}{R_x}$. So $\tan\theta=\frac{5\sqrt{3}-4}{11}\approx\frac{5\times1.732 - 4}{11}=\frac{8.66 - 4}{11}=\frac{4.66}{11}\approx0.424$. Then $\theta=\arctan(0.424)\approx22.9^{\circ}$.

Answer:

The magnitude of $\vec{A}+\vec{B}$ is approximately $11.94$ and the direction angle is approximately $22.9^{\circ}$. In polar form, it is $(11.94,22.9^{\circ})$