Definition, analytical expression and properties of scalar product

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

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

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:

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

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

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

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, $$\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|\cos(\widehat{uv})=1\cdot3\cdot\dfrac{\sqrt{3}}{2}=3\dfrac{\sqrt{3}}{2}$$