1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @FriedrichPhillip How do you know to do that? I understand what is to be done but I don't understand why. $\endgroup$ – Dunja Elez Mar 3 '16 at 1:43
  • $\begingroup$ 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 $a = |x|$. $\endgroup$ – Friedrich Philipp Mar 3 '16 at 1:49
0
$\begingroup$

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 $\sqrt{4^2+6^2} = 2\sqrt{13}$. 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 $$ \overrightarrow{c} = \begin{pmatrix} \sqrt{13} \\ \sqrt{39} \end{pmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.