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# 数学代写|线性代数代写Linear algebra代考|Angle between two vectors

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## 数学代写|线性代数代写Linear algebra代考|Angle between two vectors

Work is defined as the product of the force applied in a particular direction and the distance it moves in that direction.

A real-world example of this is to imagine a rope tied to an object, perhaps a barge. You need to move the barge by pulling the rope (Fig. 2.12).

If you could stand directly behind the barge and push, then all your force would be in the direction of movement.
In this case, $\mathbf{F}$ and $\mathbf{d}$ are parallel and the angle between them is $0^{\circ}$ so we have
$$\text { Work done }=|\mathbf{F}| \cos \left(0^{\circ}\right) \times|\mathbf{d}|=|\mathbf{F}| \times|\mathbf{d}| \quad \text { [because } \cos \left(0^{\circ}\right)=1 \text { ] }$$
This is the least possible force used to push the object because we are pushing in the same direction as we would like the object to move.

If you push the barge in a direction perpendicular (orthogonal) to the canal, it would not move forward at all, so you would do no work because
$$\text { Work done }=|\mathbf{F}| \cos \left(90^{\circ}\right) \times|\mathbf{d}|=0 \quad\left[\text { because } \cos \left(90^{\circ}\right)=0\right]$$
The actual amount of work done in moving the barge along the canal is a value somewhere between these two possibilities, and is given by the angle the rope makes with the direction of the canal.

## 数学代写|线性代数代写Linear algebra代考|Inequalities

Next, we prove some inequalities in relation to the dot product and norm (length) of vectors.
Cauchy-Schwarz inequality (2.14). Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\mathbb{R}^n$ then
$$|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|$$

For the above Example 2.6 we had
$$|\mathbf{u} \cdot \mathbf{v}| \underset{\text { By part (i) }}{\equiv} 9 \leq 21.02 \underset{\text { By part (ii) }}{\equiv}|\mathbf{u}||\mathbf{v}|$$
Proof.
How can we prove this inequality for any vectors in $\mathbb{R}^n$ ? If the vectors $\mathbf{u}$ and $\mathbf{v}$ are non-zero then we can use the above formula:
$$\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos (\theta)$$
Taking the modulus of both sides we have
$$|\mathbf{u} \cdot \mathbf{v}|=||\mathbf{u}||\mathbf{v}| \cos (\theta)|$$
The lengths are positive or zero, $|\mathbf{u}| \geq 0$ and $|\mathbf{v}| \geq 0$, therefore the modulus of these is just $|\mathbf{u}|$ and $|\mathbf{v}|$ respectively.
Why?
Because if $x \geq 0$ then $|x|=x$. Hence we have $|\mathbf{u} \cdot \mathbf{v}|=|\mathbf{u}||\mathbf{v}||\cos (\theta)|$.

## 数学代写|线性代数代写Linear algebra代考|Angle between two vectors

$$\text { Work done }=|\mathbf{F}| \cos \left(0^{\circ}\right) \times|\mathbf{d}|=|\mathbf{F}| \times|\mathbf{d}| \quad \text { [because } \cos \left(0^{\circ}\right)=1 \text { ] }$$

$$\text { Work done }=|\mathbf{F}| \cos \left(90^{\circ}\right) \times|\mathbf{d}|=0 \quad\left[\text { because } \cos \left(90^{\circ}\right)=0\right]$$

## 数学代写|线性代数代写Linear algebra代考|Inequalities

Cauchy-Schwarz不等式(2.14)。设$\mathbf{u}$和$\mathbf{v}$为$\mathbb{R}^n$中的向量
$$|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|$$

$$|\mathbf{u} \cdot \mathbf{v}| \underset{\text { By part (i) }}{\equiv} 9 \leq 21.02 \underset{\text { By part (ii) }}{\equiv}|\mathbf{u}||\mathbf{v}|$$

$$\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos (\theta)$$

$$|\mathbf{u} \cdot \mathbf{v}|=||\mathbf{u}||\mathbf{v}| \cos (\theta)|$$

## MATLAB代写

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