I know that the magnitude of a unit vector is $1$, and in order for two vectors to be parallel the angle between them should be $0^\circ$ or $180^\circ$. I also know that the formula normalizing a vector $v$ is $\dfrac{v}{\Vert v \Vert}$.
The correct answer is supposed to be $\dfrac{1}{\sqrt{50}}(i+7j)$, but I've found the magnitude of vector to be $\sqrt{7^2 + (-3)^2} = 59 $.
How can that be, and how does one get $i+7j$?
2 Answers
$\begingroup$Put $x=t$, then $y=7t-3$ and the parametric equation of the line is $$r(t)=(t,7t-3)=(0,-3)+t(1,7)$$ and hence the parallel vector is $u=(1,7)=i+7j$. Finally the unit vector is $$\frac{i+7j}{\sqrt{1^2+7^2}}$$
$\endgroup$ $\begingroup$You want:
$$x^2+(7x-3)^2=50x^2-42x+9=1 \implies 25x^2-21x+4=0$$
Solve this using the quadratic formula, and plug $x_1$ or $x_2$ back to the line equation to obtain $y$.
$(x_1,y(x_1))$ and $(x_2,y(x_2))$ will be two unit vectors parallel to the line.
