Problems from Applications of the scalar product

Find a vector $\vec{v}$ which is orthogonal (perpendicular) to the vector $\vec{u} = (2, - 4)$ which is orthogonal (perpendicular) to the vector $3$ .

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Development:

We want to find a vector $\vec{v} = (v_{1}, v_{2})$ such that its norm is $3$ , that is $| \vec{v} | = \sqrt{v_{1}^{2} + v_{2}^{2}} = 3 \Rightarrow | \vec{v} | = v_{1}^{2} + v_{2}^{2} = 9$

and that $\vec{u} \cdot \vec{v} = 0$ (we impose perpendicularity): $u_{1} v_{1} + u_{2} v_{2} = 0 \Rightarrow 2 v_{1} + (- 4) v_{2} = 0 \Rightarrow v_{1} = 2 v_{2}$

By substituting $v_{1} = 2 v_{2}$ in the first equality, we obtain: $4 v_{2}^{2} + v_{2}^{2} = 5 v_{2}^{2} = 9 \Rightarrow v_{2}^{2} = \frac{9}{5} \Rightarrow v_{2} = \frac{3}{\sqrt{5}}, v_{1} = \frac{6}{\sqrt{5}}$

Solution:

$v_{2} = \frac{3}{\sqrt{5}}$ and $v_{1} = \frac{6}{\sqrt{5}}$

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Consider the vectors $\vec{u} = (x, 2)$ and $\vec{v} = (3, 1)$ . Determine the value of $x$ so that the vectors $\vec{u}$ and $\vec{v}$ are:

Parallels.
Perpendicular.
Form an angle of $45^{\circ}$ .

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Development:

If $\vec{u}$ and $\vec{v}$ are parallel then its coordinates have to be proportional, so: $\frac{x}{3} = \frac{2}{1} \Rightarrow x = 6$
If $\vec{u}$ and $\vec{v}$ are perpendicular then its scalar product will have to be equal to zero: $\vec{u} \cdot \vec{v} = 3 x + 2 = 0 \Rightarrow x = - \frac{2}{3}$
If $\vec{u}$ and $\vec{v}$ form an angle of $45^{\circ}$ , it must be the case that: $\cos \hat{u v} = \frac{\sqrt{2}}{2}$ And for the formula of the scalar product $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos \hat{u v}$ , by substituting the cosine we obtain: $\frac{\sqrt{2}}{2} = \cos (\hat{u v}) = \frac{\vec{u} \cdot \vec{v}}{| \vec{u} | | \vec{v} |} = \frac{3 x + 2}{\sqrt{x^{2} + 4} \sqrt{10}}$ so: $\sqrt{x^{2} + 4} \cdot \sqrt{20} = 6 x + 4$ raising to the square every side of the equality: $\begin{array}{rcl} 20 (x^{2} + 4) & = & (6 x + 4)^{2} \\ 20 x^{2} + 80 & = & 36 x^{2} + 48 x + 16 \end{array}$ $16 x^{2} + 48 x - 64 = 0$ EWe need to solve the equation of second degree, which has as its solutions: $x = - 4$ and $x = 1$ . We verify if the solutions are valid.

The first value: $\vec{u} = (- 4, 2)$ , $\vec{v} = (3, 1)$ $\cos (\hat{u v}) = \frac{- 12 + 2}{\sqrt{20} \cdot \sqrt{10}} = \frac{- 10}{\sqrt{200}} = \frac{- 1}{\sqrt{2}}$ This is not valid.

The second value: $\vec{u} = (1, 2)$ , $\vec{v} = (3, 1)$ $\cos (\hat{u v}) = \frac{3 + 2}{\sqrt{5} \cdot \sqrt{10}} = \frac{5}{\sqrt{50}} = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ Yes, this is valid.

The fact that one of the solutions is not valid is the result of squaring both sides of the equation. Whenever we do this trick we always need to verify our solutions.

Solution:

$x = 6$
$x = - \frac{2}{3}$
$x = 1$

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

$\vec{u} = (1, - 2)$ , $\vec{v} = (3, 2)$
$| \vec{u} | = 3$ , $| \vec{v} | = 4$ , $ang (\vec{u}, \vec{v}) = 60^{\circ}$
Determine the angle that the vectors form $\vec{u} = (- 1, 2)$ and $\vec{v} = (3, 0)$ .

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Development:

We use the analytic expression of the scalar product: $\vec{u} \cdot \vec{v} = u_{1} v_{1} + u_{2} v_{2} = 1 \cdot 3 + (- 2) \cdot 2 = - 1$
We use the formula of the scalar product $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos (\hat{u v})$ , so that we obtain: $\vec{u} \cdot \vec{v} = 3 \cdot 4 \cdot \cos (60^{\circ}) = 12 \cdot \frac{1}{2} = 6$
We will use the formula of the scalar product $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos (\hat{u v})$ , and in order to apply it we need to calculate the norm of the vectors, so: $| \vec{u} | = \sqrt{(- 1)^{2} + 2^{2}} = \sqrt{5} | \vec{v} | = \sqrt{3^{2}} = 3$ Then we obtain: $ang (\vec{u}, \vec{v}) = \arccos (\frac{- 1 \cdot 3 + 2 \cdot 2}{\sqrt{5} \cdot 3}) = \arccos (\frac{- 3}{3 \sqrt{5}})$ $= \arccos (\frac{- 1}{\sqrt{5}}) = 116^{\circ} 33^{'} 54^{″}$

Solution:

$- 1$
$6$
$ang (\vec{u}, \vec{v}) = 116^{\circ} 33^{'} 54^{″}$

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