How do I find the intersect between the lines and sketch the lines $t\vec{a}+\vec{b}$ and $s\vec{c}+\vec{d}$?

Question

When $\vec{a}=(1,1), \vec{b}=(2,-1), \vec{c}=(2,0),$ and $\vec{d}=(1,-3)$ How do I sketch this?

Answer 1

These lines will intersect at the values $t^\ast$ and $s^\ast$ satisfying $t^\ast\mathbf{a} + \mathbf{b} = s^\ast \mathbf{c} +\mathbf{d}$. In $\mathbb{R}^2$ this can be written as the system of equations $$ t^\ast a_0 + b_0 = s^\ast c_0 + d_0\\ t^\ast a_1 + b_1 = s^\ast c_1 + d_1 $$ where I'm using the convention $\mathbf{x} = (x_0,x_1)$. You now have two equations with two unknowns.

As for sketching a graph, if you have the equation $t\, \mathbf{m} + \mathbf{b}$, then the line passes through $\mathbf{b}$ (setting $t = 0$) and heads "in the direction" of $\mathbf{m}$. (If you draw the vector $\mathbf{m}$ emanating from the point $\mathbf{b}$, that will look like a line segment of the full line).

Answer 2

To sketch it, mark the point $b$ on your page. Then draw a vector from $b$ in the direction of $a.$

Now make a line in this direction through $b.$

Do the same thing creating a line through d in the direction of $c.$

There are many ways to solve this. As others have point out how to evaluate the parameters. Another way would be to derive the standard equations of the lines.

$x-y =3\\ y = -3$