Given that the two vectors 8a -b and 4a +3b are perpendicular and that |a |= 2|b |, determine the angle between a and b.

Question

I understand you have to use the dot product rule, but I'm not sure where to take it from there.

Answer 1

This is a bit of a sketchy problem. One of the properties of the dot product is that it's linear; scaling either vector scales the whole thing, so $(a\vec{v})\cdot\vec{w}=\vec{v}\cdot(a\vec{w})=a(\vec{v}\cdot \vec{w})$. Also, the projected components add, so $\vec{u}\cdot(\vec{w}+\vec{v})=(\vec{u}\cdot\vec{w})+(\vec{u}\cdot\vec{v})$. Of course, the history of the dot product is a lot more than just linearity. Also, by the Pythagorean theorem $|\vec{v}|=\sqrt{\vec{v}\cdot\vec{v}}$. Alternatively, $\vec{v}\cdot\vec{v}=|\vec{v}|^2$, the square of the magnitude.

So if $|\vec{a}|=2|\vec{b}|$, we can rewrite this as $|\vec{a}|^2=(2|\vec{b}|)^2$, $|\vec{a}|^2=4|\vec{b}|^2$, $\vec{a}\cdot\vec{a}=4\vec{b}\cdot\vec{b}$

Anyways plugging this into the expansion of $(8a-b)\cdot(4a+3b)$, which is $(8a)\cdot(4a)+(8a)\cdot(3b)+(-b)\cdot(4a)+(-b)\cdot(3b)=32(a\cdot a)+20(a\cdot b)-3(b\cdot b)=0$

$a\cdot b$ (which you'll need to solve for the cosine, since $\cos(\theta)=\frac{a\cdot b}{|a||b|}$) equals therefore

$\frac{3(b\cdot b)-32(a\cdot a)}{20}$

$=\frac{3(b\cdot b)-32(4b\cdot b)}{20}$

$=\frac{3(b\cdot b)-128(b\cdot b)}{20}$

$=\frac{-125(b\cdot b)}{20}$

$=\frac{-25(b\cdot b)}{4}$ (diving the top and bottom by 5)

and $\cos(\theta)=\frac{a\cdot b}{|a||b|}$ which also equals $\frac{\left(\frac{-25}{4}(b\cdot b)\right)}{(2|b|)|b|}=\frac{-25}{8}$, more or less.

However as $25>8$, and cosine is always a ratio between 0 and $\pm1$, you can never have a cosine that reaches this number - not with real numbers anyways.

On desmos, I've plotted this problem. Dragging the (draggable) dot, the red shape is the set of all possible positions of tips of vectors $\vec{b}$ that satisfy $(8a-b)\cdot(4a+3b)=0$ for a given vector $\vec{a}$, the draggable point. The black circle is to make clear the magnitude of $\vec{a}$, which does seem to always be less than $|\vec{b}|$, however the problem suggests that, since $|\vec{a}|=2|\vec{b}|$, the magnitude of $\vec{a}$ is always bigger (in fact, lmao, twice as big) as $|\vec{b}|$.

If you got this problem from a textbook, these things can happen from time to time to time. Of course you can just solve $\cos^{-1}\left(\frac{-25}{8}\right)$ over the complexes, but that's probably not what your teacher expects you to do. These things happen all the time, someone slips up and makes an impossible problem. For example, I remember once in school we given a non-right triangle to solve. If you just blindly solved it using the law of sines, you happened to get a perfectly reasonable answer. But if you tried it with the law of cosines, you actually got an imaginary number. That's because the person who made the problem had set up some sketchy impossible triangle, where the two smaller side lengths didn't add up to at least the longer side length. But yeah, it's an interesting question.

So yeah, I wouldn't have answered this if it weren't for Anurag A's comment, "The above hint will work in general. But if I were you, I would want to make sure that the data given is correct." When in doubt, if you've checked your work loads of times, you can think about and dig deeper into your answer to confirm that you're right and the problem isn't. Or you copied it wrong. But still, I think that's the amazing thing about math - that you can change one little thing and it changes the whole interpretation of the problem. But, if you did copy it wrong, then it's ok as long you still know how to solve the problem - after all, math is more than just copying things! The important part is figuring out how to solve the problem, and understanding the why of stuff. The rest is just computation. I'm sorry I couldn't help more.