Problems from Definition, analytical expression and properties of scalar product

Determine the scalar product of $$\vec{u}$$ and $$\vec{v}$$:

$$\vec{u}=(1,2)$$, $$\ \vec{v}=(3,-2)$$
$$\vec{u}=(1,0)$$, $$\ \vec{v}=(0,-2)$$
$$\vec{u}=(-1,2)$$, $$\ \vec{v}=(3,0)$$

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Development:

We use the analytical expression of the scalar product: $$\vec{u}\cdot\vec{v}= u_1 v_1+u_ 2 v_2$$.

$$\vec{u}\cdot\vec{v}= u_1 v_1+u_ 2 v_2= 1\cdot3+(-2)\cdot 2=-1$$
$$\vec{u}\cdot\vec{v}= u_1 v_1+u_ 2 v_2=1\cdot0+0\cdot(-2)=0$$
$$\vec{u}\cdot\vec{v}= u_1 v_1+u_ 2 v_2=(-1)\cdot3+2\cdot0=-3$$

Solution:

$$\vec{u}\cdot\vec{v}=-1$$
$$\vec{u}\cdot\vec{v}= 0$$
$$\vec{u}\cdot\vec{v}= -3$$

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Determine the scalar product of $$\vec{u}$$ and $$\vec{v}$$ knowing that:

$$|\vec{u}|=3$$, $$\ |\vec{v}|=2$$, $$\text{ang}(\widehat{uv})=60^\circ$$
$$|\vec{u}|=5$$, $$\ |\vec{v}|=2$$, $$\cos(\widehat{uv})=\dfrac{1}{2}$$
$$|\vec{u}|=1$$, $$\ |\vec{v}|=3$$, $$\text{ang}(\widehat{uv})=30^\circ$$

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Development:

We use the definition of the scalar product: $$\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|\cos(\widehat{uv})$$

In this case if $$|\vec{u}|=3$$, $$|\vec{v}|=2$$ and $$\text{ang}(\widehat{uv})=60^\circ$$ the cosine of $$60^\circ$$ is $$\dfrac{1}{2}$$. And so, $$$ \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|\cos(\widehat{uv})=3\cdot2\cdot\dfrac{1}{2}=3$$$
In this case if $$|\vec{u}|=5$$, $$|\vec{v}|=2$$ and $$\cos(\widehat{uv})=\dfrac{1}{2}$$ then: $$$ \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|\cos(\widehat{uv})=5\cdot2\cdot\dfrac{1}{2}=5$$$
In this case if $$|\vec{u}|=1$$, $$|\vec{v}|=3$$ and $$\text{ang}(\widehat{uv})=30^\circ$$ the cosine $$30^\circ$$ is $$\dfrac{\sqrt{3}}{2}$$. And so, $$$ \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|\cos(\widehat{uv})=1\cdot3\cdot\dfrac{\sqrt{3}}{2}=3\dfrac{\sqrt{3}}{2}$$$

Solution:

$$ \vec{u}\cdot\vec{v}= 3$$
$$ \vec{u}\cdot\vec{v}=5$$
$$ \vec{u}\cdot\vec{v}= 3\dfrac{\sqrt{3}}{2}$$

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Find a vector $$\vec{v}$$ which is orthogonal (perpendicular) to the vector $$\vec{u}=(2,-4)$$.

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Development:

We want to find a $$\vec{v}=(v_1,v_2)$$ such that $$\vec{u}\cdot\vec{v}= u_1 v_1+u_ 2 v_2=0$$, since we know they must be perpendicular. Therefore, we have: $$$ \vec{u}\cdot\vec{v}=2\cdot v_1+(-4)\cdot v_2=0 \Rightarrow v_1=-2v_2$$$

So we know that we are looking for a vector such that its first component is equal to $$-2$$ times the second component. For example, $$\vec{v}=(v_1,v_2)=(-2,1)$$. More generally, if we chose any value we want for $$v_1$$ we can obtain $$v_2$$. Other examples would be:

$$\vec{v}=(v_1,v_2)=(-4,2)$$

$$\vec{v}=(v_1,v_2)=(-6,3)$$

$$\vec{v}=(v_1,v_2)=(-1,\dfrac{1}{2})$$

Solution:

Any vector such as its first component is equal to $$-2$$ times the second component.

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