Question

1. find a + b and a - b for the following cases using the sine and cosine laws (find the magnitude and direction for each case), where the magnitudes of vectors a and b are 5 and 2 inches, respectively. (a) a + b.

Explanation:

Step1: Use the law of cosines for magnitude of $\vec{a}+\vec{b}$

Let $|\vec{a}| = 5$, $|\vec{b}|=2$ and $\theta = 45^{\circ}$. The magnitude of $\vec{a}+\vec{b}$ is given by $|\vec{a}+\vec{b}|=\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}+2|\vec{a}||\vec{b}|\cos\theta}$.
\[

$$\begin{align*} |\vec{a}+\vec{b}|&=\sqrt{5^{2}+2^{2}+2\times5\times2\times\cos45^{\circ}}\\ &=\sqrt{25 + 4+20\times\frac{\sqrt{2}}{2}}\\ &=\sqrt{29 + 10\sqrt{2}}\\ &\approx\sqrt{29+10\times1.414}\\ &=\sqrt{29 + 14.14}\\ &=\sqrt{43.14}\\ &\approx6.57 \end{align*}$$

\]

Step2: Use the law of sines to find the direction of $\vec{a}+\vec{b}$

Let the angle of $\vec{a}+\vec{b}$ with $\vec{a}$ be $\alpha$. By the law of sines, $\frac{|\vec{b}|\sin\theta}{|\vec{a}+\vec{b}|}=\sin\alpha$.
\[

$$\begin{align*} \sin\alpha&=\frac{2\times\sin45^{\circ}}{\sqrt{29 + 10\sqrt{2}}}\\ &=\frac{2\times\frac{\sqrt{2}}{2}}{\sqrt{29 + 10\sqrt{2}}}\\ &=\frac{\sqrt{2}}{\sqrt{29 + 10\sqrt{2}}}\\ &\approx\frac{1.414}{\sqrt{43.14}}\\ &\approx\frac{1.414}{6.57}\\ &\approx0.215 \end{align*}$$

\]
$\alpha\approx12.4^{\circ}$

Step3: Use the law of cosines for magnitude of $\vec{a}-\vec{b}$

The magnitude of $\vec{a}-\vec{b}$ is given by $|\vec{a}-\vec{b}|=\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}||\vec{b}|\cos\theta}$.
\[

$$\begin{align*} |\vec{a}-\vec{b}|&=\sqrt{5^{2}+2^{2}-2\times5\times2\times\cos45^{\circ}}\\ &=\sqrt{25 + 4-20\times\frac{\sqrt{2}}{2}}\\ &=\sqrt{29 - 10\sqrt{2}}\\ &\approx\sqrt{29-10\times1.414}\\ &=\sqrt{29 - 14.14}\\ &=\sqrt{14.86}\\ &\approx3.85 \end{align*}$$

\]

Step4: Use the law of sines to find the direction of $\vec{a}-\vec{b}$

Let the angle of $\vec{a}-\vec{b}$ with $\vec{a}$ be $\beta$. By the law of sines, $\frac{|\vec{b}|\sin\theta}{|\vec{a}-\vec{b}|}=\sin\beta$.
\[

$$\begin{align*} \sin\beta&=\frac{2\times\sin45^{\circ}}{\sqrt{29 - 10\sqrt{2}}}\\ &=\frac{2\times\frac{\sqrt{2}}{2}}{\sqrt{29 - 10\sqrt{2}}}\\ &=\frac{\sqrt{2}}{\sqrt{29 - 10\sqrt{2}}}\\ &\approx\frac{1.414}{\sqrt{14.86}}\\ &\approx\frac{1.414}{3.85}\\ &\approx0.367 \end{align*}$$

\]
$\beta\approx21.5^{\circ}$

Answer:

Magnitude of $\vec{a}+\vec{b}\approx6.57$, direction of $\vec{a}+\vec{b}$ with $\vec{a}\approx12.4^{\circ}$, magnitude of $\vec{a}-\vec{b}\approx3.85$, direction of $\vec{a}-\vec{b}$ with $\vec{a}\approx21.5^{\circ}$