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: $\vec{u} \cdot \vec{v} = u_{1} v_{1} + u_{2} v_{2}$ .

$\vec{u} \cdot \vec{v} = u_{1} v_{1} + u_{2} v_{2} = 1 \cdot 3 + (- 2) \cdot 2 = - 1$
$\vec{u} \cdot \vec{v} = u_{1} v_{1} + u_{2} v_{2} = 1 \cdot 0 + 0 \cdot (- 2) = 0$
$\vec{u} \cdot \vec{v} = u_{1} v_{1} + u_{2} v_{2} = (- 1) \cdot 3 + 2 \cdot 0 = - 3$

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:

$| \vec{u} | = 3$ , $| \vec{v} | = 2$ , $ang (\hat{u v}) = 60^{\circ}$
$| \vec{u} | = 5$ , $| \vec{v} | = 2$ , $\cos (\hat{u v}) = \frac{1}{2}$
$| \vec{u} | = 1$ , $| \vec{v} | = 3$ , $ang (\hat{u v}) = 30^{\circ}$

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Development:

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

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

Solution:

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

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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: $\vec{u} \cdot \vec{v} = 2 \cdot v_{1} + (- 4) \cdot v_{2} = 0 \Rightarrow v_{1} = - 2 v_{2}$

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:

$\vec{v} = (v_{1}, v_{2}) = (- 4, 2)$

$\vec{v} = (v_{1}, v_{2}) = (- 6, 3)$

$\vec{v} = (v_{1}, v_{2}) = (- 1, \frac{1}{2})$

Solution:

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

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