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Calculate the norms of the following vector:-

1- $x=\left(\begin{array}{c}{1} \\ {-1}\end{array}\right)$

$\Rightarrow|x|=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$

2- $x=\left(\begin{array}{c}{-2} \\ {2} \\ {1}\end{array}\right)$

$\Rightarrow|x|=\sqrt{(-2)^{2}+(2)^{2}+(1)^{2}}=\sqrt{9}=3$

Let $x \in R^{n}$ and $\lambda \in R .$ Show that the following holds :-
$\|\lambda \cdot x\|=|\lambda| \cdot\|x\|$

LHS= $\Rightarrow\|\lambda x\|=\sqrt{\left(\lambda_{1} x_{1}\right)^{2}+\left(\lambda_{2} x_{2}\right)^{2}+\cdots \cdot\left(\lambda_{n} x_{n}\right)^{2}}$

Let $x_{n}=\left[\begin{array}{c}{x_{1}} \\ {x_{2}} \\ {\vdots} \\ {x_{n}}\end{array}\right]$

$\lambda x_{n}=\left[\begin{array}{c}{\lambda_{1} x_{1}} \\ {\lambda_{2} x_{2}} \\ {\vdots} \\ {\lambda_{n} x_{n}}\end{array}\right]$

$\lambda x_{n}=\left[\begin{array}{ccc}{\lambda} {x_{1}} \\ {\lambda} {x_{2}} \\ {} {\vdots} \\ {\lambda} {x_{n}}\end{array}\right]$

$\|\lambda x\|=\sqrt{\left(\lambda x_{1}\right)^{2}+\left(\lambda x_{2}\right)^{2}+\cdots \cdot\left(\lambda x_{n}\right)^{2}}$

$=\sqrt{\lambda^{2} x_{1}^{2}+\lambda^{2} x_{2}^{2}+\cdots \cdot \quad \lambda^{2} x_{n}^{2}}$

$=\sqrt{\lambda^{2}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)}$

$=|\lambda| \cdot \sqrt{x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}}$

$=|\lambda| \cdot\left\|x_{n}\right\|=$RHS

Exercise 2 Find the distance between each pair of the following pairs of vectors :-
And which of the above pairs represents orthogonal vectors? Justify your answer.
If they are not orthogonal, find the angle between these vectors.

1- $\left(\begin{array}{c}{1} \\ {-2}\end{array}\right)$ and $\left(\begin{array}{l}{3} \\ {0}\end{array}\right)$

$\Rightarrow d(u, v)=\|v-u\| \Rightarrow v-u=\left(\begin{array}{c}{3} \\ {0}\end{array}\right)-\left(\begin{array}{c}{1} \\ {-2}\end{array}\right)=\left(\begin{array}{l}{2} \\ {2}\end{array}\right) \Rightarrow|v-u|=\sqrt{2^{2}+2^{2}}=\sqrt{8}$

$u \cdot v=(1 \times 3)+(-2 \times 0)=3+0=3 \neq 0$ Not orthogonal

$\cos (\theta)=\frac{u \cdot v}{|u| | v |}=\frac{3}{\sqrt{1^{2}+(-2)^{2}} \sqrt{3^{2}}}$

$\cos (\theta)=\frac{3}{3 \sqrt{5}}=1 / \sqrt{5}$

$\theta=\cos ^{-1}(1 / \sqrt{5})$

2- $\left(\begin{array}{c}{1} \\ {-2} \\ {3}\end{array}\right) \text { and }\left(\begin{array}{l}{4} \\ {5} \\ {2}\end{array}\right)$

$\Rightarrow d(u, v)=\|v-u\|$

$v-u=\left(\begin{array}{c}{4} \\ {5} \\ {2}\end{array}\right)-\left(\begin{array}{c}{1} \\ {-2} \\ {3}\end{array}\right)=\left(\begin{array}{c}{1} \\ {7} \\ {-1}\end{array}\right) \Rightarrow\|v-u\|=\sqrt{3^{2}+7^{2}+(-1)^{2}}=\sqrt {59}$

$u \cdot v=(1 \times 4)+(-2 \times 5)+(3 \times 2)$

$=4+(-10)+6=0 \Rightarrow$ orthogonal

Exercise 3 Normalize the following vectors:

1- $x=\left(\begin{array}{l}{3} \\ {0} \\ {4}\end{array}\right)$

$\Rightarrow|x|=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5$

$\hat{x}=\frac{\vec{x}}{| | x| |}=\frac{1}{\|x\|} \cdot \vec{x}=\frac{3 i+4 k}{5}=3 / 5 i+0 j+4 / 5 k \Rightarrow \hat{x}=\left(\begin{array}{c}{3 / 5} \\ 0 \\ {4 / 5}\end{array}\right)$

2- $x=\left(\begin{array}{l}{1} \\ {2} \\ {0} \\ {0}\end{array}\right)$

$\Rightarrow|x|=\sqrt{1^{2}+2^{2}}=\sqrt{5}$

$\hat x=\frac{1}{\sqrt{5}}\left(\begin{array}{l}{1} \\ {2} \\ {0} \\ {0}\end{array}\right)=\left(\begin{array}{c}{1 / \sqrt{5}} \\ {2 / \sqrt{5}} \\ {0} \\ {0}\end{array}\right)$