Problems from Fundamental concepts of vectors

Development:

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: $$\begin{array}{l} |(1,3)|=\sqrt{1^2+3^2}=\sqrt{10} \\ |(-1,2)|=\sqrt{(-1)^2+2^2}=\sqrt{5} \\ |(-7,0)|=\sqrt{(-7)^2+0^2}=\sqrt{49}=7 \end{array} $$

Solution:

1. $$(1,3)$$
2. $$(-1,2)$$
3. $$(-7,0)$$
4. $$\sqrt{10}$$, $$\sqrt{5}$$, $$7$$

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Development:

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: $$|(8,-2)|=\sqrt{8^2+(-2)^2}=\sqrt{68}=2\sqrt{17}$$.

Solution:

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

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Development:

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: $$$ |\overrightarrow{AB}|=|(-2,5)|=\sqrt{(-2)^2+5^2}=\sqrt{4+25}=\sqrt{29}$$$

Solution:

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

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