Find a vector equation and parametric equation of the line in $\mathbb{R}^3$ that passes through the point $(1,2,-3)$ and is parallel to the vector $u=(4,-5,1)$. Find two points on the line that are different from the point $(1,2,-3)$
Here is what I've done so far:
$(x,y,z)=(x_0,y_0,z_0)+(a,b,c)t =(1,2,-3) + (4,-5,1)t$
$x=1+4t$, $y=2-5t$, $z=-3+t$ (then I don't know how to solve it...should I find "t" first? if yes, how can I solve it)
thanks for helping... :)
You don't “solve” this in the way that you compute a unique and well-defined solution. Instead, you choose values for parameter $t$ to obtain points on the line. For a given choice of $t$ (which one?) this will be the “forbidden” point $(1,2,-3)$, for all other values it will be a different point. As long as there are no further restrictions than those you quoted, any choice except the single forbidden one will lead to a valid point, and any two valid points form a valid answer. So you can choose the values you like best, be it small values, or some such that the resulting points resemble your phone number, or some encoding of the birthday of your loved one, or simply the first two you could think of.