Fundamental concepts of vectors

Determine the values of $x$ and $y$ so that the following identities hold:

  1. $(x,y)-3(2,5)=(4,1)$
  2. $2(1,x)+3(y,2)=(8,-2)$
  3. Compute the norm of the vectors $(4,1)$ and $(8,-2)$.
  1. $(x,y)=(4,1)+3(2,5)=(4,1)+(6,15)=(10,16)$ and we have $x=10$, $y=16$.
  2. $2(1,x)+3(y,2)=(2,2x)+(3y,6)=(2+3y,2x+6)=(8,-2)$. $$\left. \begin{array}{l} 2+3y=8 \\ 2x+6=-2 \end{array} \right\} \Rightarrow x=-4, \ y=2$$
  3. We apply the formula $|\vec{u}|=\sqrt{u_1^2+u_2^2}$ to our vectors.

For the first one we obtain: $|(4,1)|=\sqrt{4^2+1^2}=\sqrt{17}$.

And for the second one: $|(8,-2)|=\sqrt{8^2+(-2)^2}=\sqrt{68}=2\sqrt{17}$.

  1. $(4,2)$
  2. $(10,16)$
  3. $\sqrt{17}$ and $2\sqrt{17}$

Compute the components of the vectors whose origin and end are given by the following points:

  1. Origin $(-1,3)$, end $(0,6)$.
  2. Origin $(2,-1)$, end $(1,1)$.
  3. Origin $(5,1)$, end $(-2,1)$.
  4. Compute the norm of the vectors obtained in the previous points.

We subtract the components of the ending point from those of the origin.

  1. $(0,6)-(-1,3)=(1,3)$.
  2. $(1,1)-(2,-1)=(-1,2)$.
  3. $(-2,1)-(5,1)=(-7,0)$.
  4. We use the formula $|\vec{u}|=\sqrt{u_1^2+u_2^2}$, to obtain: $\begin{array}{l} |(1,3)|=\sqrt{1^2+3^2}=\sqrt{10} \ |(-1,2)|=\sqrt{(-1)^2+2^2}=\sqrt{5} \ |(-7,0)|=\sqrt{(-7)^2+0^2}=\sqrt{49}=7 \end{array} $
  1. $(1,3)$
  2. $(-1,2)$
  3. $(-7,0)$
  4. $\sqrt{10}$, $\sqrt{5}$, $7$

Compute the components of the ending point of the vector $\overrightarrow{AB}=(-2,5)$ if we know that the origin $A$ is $(1,1)$. And find the norm of the vector $\overrightarrow{AB}$.

Since we have obtained the components of the vector by subtracting the components of the end from those of the origin, we have: $$ (b_1,b_2)=(-2,5)+(1,1)=(-1,6)$$ The norm of the vector $\overrightarrow{AB}$ by using the formula $|\vec{u}|=\sqrt{u_1^2+u_2^2}$ will be: $$ |\overrightarrow{AB}|=|(-2,5)|=\sqrt{(-2)^2+5^2}=\sqrt{4+25}=\sqrt{29}$$

$(-1,6)$, $\sqrt{29}$.

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