# Unit Length Vector and Dot Product

Suppose $\vec a$ = [4, 6] and $\vec b$ = [1, 2]. Determine:

a) A vector with unit length in the opposite direction to $\vec b$

For this question I understand I would have to use the $\vec a$ = k ($\vec b$) equation since we are a talking about opposite direction which I would consider collinear and from there using the magnitude equation to equal $1$ since that is the unit length and I would substitute the result of $\vec a$ = k ($\vec b$) like so.. $$1=\sqrt (k^2+2k^2)$$ $$1=5k^2$$ $${ 1 \over\sqrt 5} = k$$ But now I have no idea what to do next because the final answer comes to [$-\sqrt 5 \over 5$,$-2\sqrt 5 \over 5$]. Have I done everything correct so far? What do I need to do next?

b) The components of a vector with the same magnitude as $\vec a$ making an angle of $60^\circ$ with the positive x-axis.

I have no idea how to do this question but I feel like I would have to use the dot product for it

This is really simple, Dunja. First, normalize the vector $b$ by dividing each of its coordinates by the length of $b$ (which is $\sqrt 5$, here). By this, you shrink the vector to length one. Then put a minus in front of this normalized vector to make it opposite direction. Done.

• @FriedrichPhillip How do you know to do that? I understand what is to be done but I don't understand why. Mar 3, 2016 at 1:43
• Well, let $|x|$ denote the length of a vector $x$. It is easy to prove that for a positive number $a$ we have $|ax| = a\cdot|x|$. Now, let $a = |x|$. Then the length of the vector $(1/a)x$ is $|(1/a)x| = (1/a)|x| = 1$ since $a = |x|$. Mar 3, 2016 at 1:49

Recall that the formula for the angle between two vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ is $$\cos(\theta) = \frac{\overrightarrow{a} \bullet \overrightarrow{b}}{||\overrightarrow{a}||\cdot||\overrightarrow{b}||}$$
Now, let's define our two vectors $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$. We know that $$\overrightarrow{a}$$ is $$\begin{pmatrix} 4 \\ 6 \end{pmatrix}$$so therefore its magnitude is $$\sqrt{4^2+6^2} = 2\sqrt{13}$$. We know that the x-axis is a horizontal line, and we can represent any vector along it as $$\begin{pmatrix} k \\ 0 \end{pmatrix}$$ where $$k \in \mathbb{R}$$. For simplicity's sake, lets let $$k = 1$$. Now, we define an arbitrary vector $$\overrightarrow{c}$$ such that $$\overrightarrow{c} = \begin{pmatrix} x \\ y \end{pmatrix}$$ Then, $$||\overrightarrow{c}|| = \sqrt{x^2+y^2}$$
We now have all the information we need to solve this problem. We want the angle between our two vectors to be $$60^\circ$$, so the LHS of our first equation becomes $$\cos(60^\circ) = \frac{1}{2}$$ $$\frac{1}{2} = \frac{\overrightarrow{c} \bullet \overrightarrow{x}}{2\sqrt{13}}$$ The dot product of $$\overrightarrow{c} \bullet \overrightarrow{x} = x$$, so our after cross-multiplying, our equation becomes $$2x = 2\sqrt{13} \Longrightarrow x = \sqrt{13}$$
Remember, we're solving for the vector $$\overrightarrow{c}$$, and so far, we only know the value of the $$x$$ component of that vector. We know that $$||\overrightarrow{c}|| = 2\sqrt{13}$$, so $$\sqrt{x^2+y^2} = 2\sqrt{13}$$ Squaring both sides:$$x^2+y^2=52 \Longrightarrow 13 + y^2 = 52$$ From there we can solve for $$y$$, leaving us with $$y = \sqrt{39}$$. Therefore, our vector $$\overrightarrow{c} = \begin{pmatrix} \sqrt{13} \\ \sqrt{39} \end{pmatrix}$$