Riemann surface
matlab

Problem 1.1
In terms of the standard basis set ${\boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k}}, \boldsymbol{a}=2 \boldsymbol{i}-\boldsymbol{j}-2 \boldsymbol{k}, \boldsymbol{b}=3 \boldsymbol{i}-4 \boldsymbol{k}$ and $\boldsymbol{c}=\boldsymbol{i}-5 \boldsymbol{j}+3 \boldsymbol{k}$
(i) Find $3 \boldsymbol{a}+2 \boldsymbol{b}-4 \boldsymbol{c}$ and $|\boldsymbol{a}-\boldsymbol{b}|^2$.
(ii) Find $|\boldsymbol{a}|,|\boldsymbol{b}|$ and $\boldsymbol{a} \cdot \boldsymbol{b}$. Deduce the angle between $\boldsymbol{a}$ and $\boldsymbol{b}$.
(iii) Find the component of $\boldsymbol{c}$ in the direction of $\boldsymbol{a}$ and in the direction of $\boldsymbol{b}$.
(iv) Find $\boldsymbol{a} \times \boldsymbol{b}, \boldsymbol{b} \times \boldsymbol{c}$ and $(\boldsymbol{a} \times \boldsymbol{b}) \times(\boldsymbol{b} \times \boldsymbol{c})$.
(v) Find $\boldsymbol{a} \cdot(\boldsymbol{b} \times \boldsymbol{c})$ and $(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}$ and verify that they are equal. Is the set ${\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}}$ right- or left-handed?
(vi) By evaluating each side, verify the identity $\boldsymbol{a} \times(\boldsymbol{b} \times \boldsymbol{c})=(\boldsymbol{a} \cdot \boldsymbol{c}) \boldsymbol{b}-(\boldsymbol{a} \cdot \boldsymbol{b}) \boldsymbol{c}$.

Solution
(i)
\begin{aligned} & 3 \boldsymbol{a}+2 \boldsymbol{b}-4 \boldsymbol{c}=3(2 \boldsymbol{i}-\boldsymbol{j}-2 \boldsymbol{k})+2(3 \boldsymbol{i}-4 \boldsymbol{k})-4(\boldsymbol{i}-5 \boldsymbol{j}+3 \boldsymbol{k}) \ & =8 i+17 \boldsymbol{j}-26 \boldsymbol{k} \text {. } \ & |\boldsymbol{a}-\boldsymbol{b}|^2=(\boldsymbol{a}-\boldsymbol{b}) \cdot(\boldsymbol{a}-\boldsymbol{b}) \ & =(-\boldsymbol{i}-\boldsymbol{j}+2 \boldsymbol{k}) \cdot(-\boldsymbol{i}-\boldsymbol{j}+2 \boldsymbol{k}) \ & =(-1)^2+(-1)^2+2^2=6 \text {. } \ & \end{aligned}
(ii)
\begin{aligned} |\boldsymbol{a}|^2 & =\boldsymbol{a} \cdot \boldsymbol{a} \ & =(2 \boldsymbol{i}-\boldsymbol{j}-2 \boldsymbol{k}) \cdot(2 \boldsymbol{i}-\boldsymbol{j}-2 \boldsymbol{k}) \ & =2^2+(-1)^2+(-2)^2=9 . \end{aligned}
Hence $|a|=3$.
\begin{aligned} |\boldsymbol{b}|^2 & =\boldsymbol{b} \cdot \boldsymbol{b} \ & =(3 \boldsymbol{i}-4 \boldsymbol{k}) \cdot(3 \boldsymbol{i}-4 \boldsymbol{k}) \ & =3^2+(-4)^2=25 . \end{aligned}
Hence $|\boldsymbol{b}|=5$.
\begin{aligned} \boldsymbol{a} \cdot \boldsymbol{b} & =(2 \boldsymbol{i}-\boldsymbol{j}-2 \boldsymbol{k}) \cdot(3 \boldsymbol{i}-4 \boldsymbol{k}) \ & =(2 \times 3)+((-1) \times 0)+((-2) \times(-4)) \ & =14 . \end{aligned}

Problem 1.3
$A B C D E F$ is a regular hexagon with centre $O$ which is also the origin of position vectors. Find the position vectors of the vertices $C, D, E, F$ in terms of the position vectors $\boldsymbol{a}, \boldsymbol{b}$ of $A$ and $B$.
FIGURE $1.2 A B C D E F$ is a regular hexagon.

Solution
(i) The position vector $c$ is represented by the line segment $\overrightarrow{O C}$ which has the same magnitude and direction as the line segment $\overrightarrow{A B}$. Hence
$$\boldsymbol{c}=\boldsymbol{b}-\boldsymbol{a}$$
(ii) The position vector $d$ is represented by the line segment $\overrightarrow{O D}$ which has the same magnitude as, but opposite direction to, the line segment $\overrightarrow{O A}$. Hence
$$d=-a$$
(iii) The position vector $e$ is represented by the line segment $\overrightarrow{O E}$ which has the same magnitude as, but opposite direction to, the line segment $\overrightarrow{O B}$. Hence
$$e=-b$$
(iv) The position vector $f$ is represented by the line segment $\overrightarrow{O F}$ which has the

same magnitude as, but opposite direction to, the line segment $\overrightarrow{A B}$. Hence
$$e=-(b-a)=\boldsymbol{a}-\boldsymbol{b} .$$

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