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

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

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

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