Problems from Applications of the scalar product

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): $$$ u_1 v_1+u_ 2 v_2=0 \Rightarrow 2v_1+(-4)v_2=0 \Rightarrow v_1=2v_2$$$

By substituting $$v_1=2v_2$$ in the first equality, we obtain: $$$ 4v_2^2+v_2^2=5v_2^2=9 \Rightarrow v_2^2=\dfrac{9}{5} \Rightarrow v_2=\dfrac{3}{\sqrt{5}}, \ v_1=\dfrac{6}{\sqrt{5}}$$$

Solution:

$$v_2=\dfrac{3}{\sqrt{5}}$$ and $$v_1=\dfrac{6}{\sqrt{5}}$$

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Development:

• If $$\vec{u}$$ and $$\vec{v}$$ are parallel then its coordinates have to be proportional, so: $$$ \dfrac{x}{3}=\dfrac{2}{1}\Rightarrow x=6$$$
• If $$\vec{u}$$ and $$\vec{v}$$ are perpendicular then its scalar product will have to be equal to zero: $$$\vec{u}\cdot\vec{v}=3x+2=0 \Rightarrow x=-\dfrac{2}{3}$$$

• If $$\vec{u}$$ and $$\vec{v}$$ form an angle of $$45^\circ$$, it must be the case that: $$$\cos{\widehat{uv}}=\dfrac{\sqrt{2}}{2}$$$ And for the formula of the scalar product $$\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos{\widehat{uv}}$$, by substituting the cosine we obtain: $$$\dfrac{\sqrt{2}}{2}=\cos(\widehat{uv})= \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}= \dfrac{3x+2}{\sqrt{x^2+4}\sqrt{10}}$$$ so: $$$\sqrt{x^2+4}\cdot\sqrt{20}=6x+4$$$ raising to the square every side of the equality: $$$ \begin{array}{rcl} 20(x^2+4) &=& (6x+4)^2 \\ 20x^2+80 &=& 36x^2+48x+16 \end{array}$$$ $$$16x^2+48x-64 =0$$$ 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: $$\vec{u}=(-4,2)$$, $$\vec{v}=(3,1)$$ $$$\cos(\widehat{uv})=\dfrac{-12+2}{\sqrt{20}\cdot\sqrt{10}}= \dfrac{-10}{\sqrt{200}}=\dfrac{-1}{\sqrt{2}}$$$ This is not valid.

  The second value: $$\vec{u}=(1,2)$$, $$\vec{v}=(3,1)$$ $$$\cos(\widehat{uv})=\dfrac{3+2}{\sqrt{5}\cdot\sqrt{10}}= \dfrac{5}{\sqrt{50}}=\dfrac{5}{5\sqrt{2}}=\dfrac{1}{\sqrt{2}}= \dfrac{\sqrt{2}}{2}$$$ 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:

• $$x=6$$
• $$x=-\dfrac{2}{3}$$
• $$x=1$$

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Development:

• We use the analytic expression of the scalar product: $$$\vec{u}\cdot\vec{v}= u_1 v_1+u_ 2 v_2=1\cdot3+(-2)\cdot2=-1$$$
• We use the formula of the scalar product $$\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|\cos(\widehat{uv})$$, so that we obtain: $$$\vec{u}\cdot\vec{v}=3\cdot4\cdot\cos(60^\circ)=12\cdot\dfrac{1}{2}=6$$$
• We will use the formula of the scalar product $$\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|\cos(\widehat{uv})$$, and in order to apply it we need to calculate the norm of the vectors, so: $$$|\vec{u}|=\sqrt{(-1)^2+2^2}=\sqrt{5} \qquad |\vec{v}|=\sqrt{3^2}=3$$$ Then we obtain: $$$\text{ang}(\vec{u},\vec{v})= \arccos\Big(\dfrac{-1\cdot3+2\cdot2}{\sqrt{5}\cdot3}\Big) =\arccos\Big(\dfrac{-3}{3\sqrt{5}}\Big)$$$ $$$=\arccos\Big( \dfrac{-1}{\sqrt{5}}\Big)=116^\circ 33' 54''$$$

Solution:

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

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