Answer
a) ${{\left\| \mathbf{u} \right\|}^{2}}={{\left\| \mathbf{v} \right\|}^{2}}+{{\left\| \mathbf{w} \right\|}^{2}}-2\left\| \mathbf{u} \right\|\left\| \mathbf{v} \right\|\cos \theta $
b) $\left\| \mathbf{u} \right\|=\sqrt{{{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}}$, ${{\left\| \mathbf{u} \right\|}^{2}}={{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}$, $\left\| \mathbf{v} \right\|=\sqrt{a_{1}^{2}+b_{1}^{2}}$, ${{\left\| \mathbf{v} \right\|}^{2}}=a_{1}^{2}+b_{1}^{2}$, $\left\| \mathbf{w} \right\|=\sqrt{a_{2}^{2}+b_{2}^{2}}$, and ${{\left\| \mathbf{w} \right\|}^{2}}=a_{2}^{2}+b_{2}^{2}$.
Work Step by Step
(a)
According to the cosine law of triangles
${{\left\| \mathbf{A} \right\|}^{2}}={{\left\| \mathbf{B} \right\|}^{2}}+{{\left\| \mathbf{C} \right\|}^{2}}-2\left\| \mathbf{B} \right\|\left\| \mathbf{C} \right\|\cos \theta $
Here, $\theta $ is the angle between the vectors $\mathbf{B}$ and $\mathbf{C}$.
From the given figure, the angle between the vectors $\mathbf{u}$ and $\mathbf{w}$ is $\theta $. Now, apply the cosine law for vector $\mathbf{u}$ to get
${{\left\| \mathbf{u} \right\|}^{2}}={{\left\| \mathbf{v} \right\|}^{2}}+{{\left\| \mathbf{w} \right\|}^{2}}-2\left\| \mathbf{u} \right\|\left\| \mathbf{v} \right\|\cos \theta $
(b)
The vector $\mathbf{u}$ is
$\mathbf{u}=\left( {{a}_{1}}-{{a}_{2}} \right)\mathbf{i}+\left( {{b}_{1}}-{{b}_{2}} \right)\mathbf{j}$
The magnitude of the vector $\mathbf{u}$ is
$\left\| \mathbf{u} \right\|=\sqrt{{{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}}$
Squaring on both sides, we get
${{\left\| \mathbf{u} \right\|}^{2}}={{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}$
The vector $\mathbf{w}$ is
$\begin{align}
& \mathbf{w}=\left( {{a}_{2}}-0 \right)\mathbf{i}+\left( {{b}_{2}}-0 \right)\mathbf{j} \\
& ={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}
\end{align}$
The magnitude of the vector $\mathbf{w}$ is
$\left\| \mathbf{w} \right\|=\sqrt{a_{2}^{2}+b_{2}^{2}}$
Squaring on both sides, we get
${{\left\| \mathbf{w} \right\|}^{2}}=a_{2}^{2}+b_{2}^{2}$
The vector $\mathbf{w}$ is
$\begin{align}
& \mathbf{v}=\left( {{a}_{1}}-0 \right)\mathbf{i}+\left( {{b}_{1}}-0 \right)\mathbf{j} \\
& ={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}
\end{align}$
The magnitude of the vector $\mathbf{v}$ is
$\left\| \mathbf{v} \right\|=\sqrt{a_{1}^{2}+b_{1}^{2}}$
Squaring on both sides, we get
${{\left\| \mathbf{v} \right\|}^{2}}=a_{1}^{2}+b_{1}^{2}$
