Csdrhrt

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..
$$1=sqrt (k^2+2k^2)$$
$$1=5k^2$$
$${ 1 oversqrt 5} = k$$
But now I have no idea what to do next because the final answer comes to [$-sqrt 5 over 5$,$-2sqrt 5 over 5$]. 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.

Since a) has already been answered, I'll go ahead and answer just b)

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 $sqrt{4^2+6^2} = 2sqrt{13}$. 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 $cos(60^circ) = frac{1}{2}$
$$frac{1}{2} = frac{overrightarrow{c} bullet overrightarrow{x}}{2sqrt{13}}$$
The dot product of $overrightarrow{c} bullet overrightarrow{x} = x$, so our after cross-multiplying, our equation becomes $2x = 2sqrt{13} Longrightarrow x = sqrt{13}$

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 $||overrightarrow{c}|| = 2sqrt{13}$, so $$sqrt{x^2+y^2} = 2sqrt{13}$$
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 $$ overrightarrow{c} = begin{pmatrix}
sqrt{13} \
sqrt{39}
end{pmatrix}$$