2. if |a| = 15cm, |b| = 22 cm, and the two vectors make an angle of 135°, find the magnitude and the angle of the following vectors. show all of your work and include a diagram in your answers.
a. -\vec{b}+\vec{a} 4 marks

Question

Accepted Answer

# Explanation:
## Step1: Find the magnitude of $-\vec{b}+\vec{a}$
Use the law of cosines $|\vec{c}|^{2}=|\vec{m}|^{2}+|\vec{n}|^{2}-2|\vec{m}||\vec{n}|\cos\theta$, where $\vec{c}=-\vec{b}+\vec{a}$, $|\vec{m}| = |\vec{a}|=15$, $|\vec{n}|=|\vec{b}| = 22$, and $\theta = 180^{\circ}- 135^{\circ}=45^{\circ}$
$$
\begin{align*}
|-\vec{b}+\vec{a}|^{2}&=15^{2}+22^{2}-2\times15\times22\times\cos45^{\circ}\
&=225 + 484-660\times\frac{\sqrt{2}}{2}\
&=709-330\sqrt{2}\
\end{align*}
$$
$$|-\vec{b}+\vec{a}|=\sqrt{709 - 330\sqrt{2}}\approx\sqrt{709-466.69}\approx\sqrt{242.31}\approx15.57$$

## Step2: Find the angle of $-\vec{b}+\vec{a}$
Use the law of sines $\frac{\sin\alpha}{|\vec{b}|}=\frac{\sin\theta}{|-\vec{b}+\vec{a}|}$, where $\alpha$ is the angle between $\vec{a}$ and $-\vec{b}+\vec{a}$
$$
\begin{align*}
\sin\alpha&=\frac{|\vec{b}|\sin\theta}{|-\vec{b}+\vec{a}|}\
&=\frac{22\times\sin45^{\circ}}{15.57}\
&=\frac{22\times\frac{\sqrt{2}}{2}}{15.57}\
&=\frac{15.56}{15.57}\approx0.999
\end{align*}
$$
$\alpha\approx87.9^{\circ}$

# Answer:
The magnitude of $-\vec{b}+\vec{a}$ is approximately $15.57$ cm and the angle (with respect to $\vec{a}$) is approximately $87.9^{\circ}$

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Question

Explanation:

Step1: Find the magnitude of \(-\vec{b}+\vec{a}\)

Step2: Find the angle of \(-\vec{b}+\vec{a}\)

Answer: