Unit Length Vector and Dot Product

$\begingroup$

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

2 Answers

$\begingroup$

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.

$\endgroup$ 2 $\begingroup$

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

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

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