dot products with vectors defined by points

Question

I have a vector defined parametrically as $\vec{v}_1 = (1,1) + t(2,1)$ and a second one defined as $\vec{v}_2 = (4,5) + t(1, -1)$ and I want to know what $\vec{v}_1 \cdot \vec{v}_2$ is, however I dont know how to apply either of the definitions of dot product that Ive found. For

$$|a||b|\cos\theta $$

I need to know both the angle between the vectors (doable, but relatively difficult) and the length of the vectors (impossible because they don't have an end.

For

$$x_1\cdot x_2 + y_1\cdot y_2$$

I don't understand how to convert a vector which is not on the origin into one that this formula accepts. It appears to presuppose that my vector starts on the origin.

Please let me know if this particular example makes the question hard.

Answer 1

A two dimensional vector has two coordinates: $(a,b)$ where $a$ says the length the vector goes to the right and $b$ is the length it goes up.

A vector can be placed anywhere, it will remain the same vector.
So, nothing prevents us to place it on the origin.

Finally, yes, the $x_1x_2+y_1y_2$ formula is to be used: noting that $v_1=(1+2t,\ 1+t)$ and $v_2=(4+t,\ 5-t)$, we get $$v_1\cdot v_2\ =\ (1+2t)(4+t)\ +\ (1+t)(5-t)\,.$$

Answer 2

you have $v_1=(1+2t,1+t)$ and $v_2=(4+t,5-t)$. Hence their inner product will be $(1+2t)(4+t)+(1+t)(5-t)=t^2+13t+9$

Answer 3

These are equation of lines. So you need to find the intersection of the lines and then calculate/make vector along each line and find angle.

1. Suppose at interaction $t = t_{1}$ and $t = t_{2}$ Solve $(1,1) + t_{1} (2,1) = (4,5) + t_{2} (1, -1)$. $t_{1} = \frac{7}{3}$ and $t_{2} = \frac{5}{3}$
2. Interaction point is $\vec{V} = (\frac{17}{3}, \frac{10}{3}) $
3. For $t_{1} = 1$, $V_{1} = (3,2)$ and $t_{2} = 1$, $V_{2} = (5,4)$
4. Now find $\frac{(V_{1} - V) \cdot (V_{2} - V)}{\| V_{1} - V \| \| V_{2} - V \|}$