Vectors Question (From Calc & Vectors class)

[ecilalikespie]

Hello, so I have a problem. I am struggling with this vectors problem. I wasn't sure where to post this but since it's from my calc class I decided to post it here. I am not sure what to do after what I have already done, and would like to know if what I have already done is correct. Thank you in advance for anyone who takes the time out of their day to help me, it is very much appreciated.

 

[Dr.Peterson]

Hello, so I have a problem. I am struggling with this vectors problem. I wasn't sure where to post this but since it's from my calc class I decided to post it here. I am not sure what to do after what I have already done, and would like to know if what I have already done is correct. Thank you in advance for anyone who takes the time out of their day to help me, it is very much appreciated.View attachment 32169View attachment 32170

Good work so far. Now just replace

in


The unknown will cancel out.

 

[pka]

View attachment 32170

Frankly I simply don' understand your work.
The problem requires finding [imath]\dfrac{\overrightarrow a \cdot\overrightarrow {b} }{\left\|\overrightarrow{a} \right\|\left\|\overrightarrow b \right\|}=\cos(\theta)[/imath] where [imath]\theta[/imath] is the angle between the vectors.
You are correct [imath]\left(3\overrightarrow {a}+2\overrightarrow {b}\right)\cdot\left(2\overrightarrow {a}-3\overrightarrow {b}\right)=0[/imath]
That gives us: [imath]6\left\|\overrightarrow{a}\right\|^2-5\overrightarrow {a}\cdot\overrightarrow {b}-6\left\|\overrightarrow{b}\right\|^2=0[/imath]
Using the given, [imath]2\left\|\overrightarrow{a}\right\|=3\left\|\overrightarrow{b}\right\|[/imath] can you find [imath]\cos(\theta)~?[/imath][imath][/imath][imath][/imath]

 

[lex]

@ecilalikespie
You made a slight error here:




Depending on your definition of 'perpendicular', this question can be considered to have no solution.


[imath]\hspace3ex (3\utilde{a} + 2\utilde{b}) \perp (2\utilde{a}-3\utilde{b}),\hspace8ex 2a=3b\\ \;\\ \Rightarrow (3\utilde{a} + 2\utilde{b}) . (2\utilde{a}-3\utilde{b})=0\\ \;\\ \Rightarrow 6a^2-6b^2-5\utilde{a}.\utilde{b}=0\\ \;\\ \Rightarrow 5ab-5\utilde{a}.\utilde{b}=0\hspace4ex \text{(since }2a=3b \Rightarrow 6a^2=9ab \text{ and }6b^2=4ab)\\ \;\\ \therefore \cos\theta=1 \hspace3ex \text{(} \theta \text{ angle between }\utilde{a}, \utilde{b}\text{)}\\ \;\\ \Rightarrow \theta=0\\ \;\\ \Rightarrow \boxed{\;\utilde{a}=\tfrac{3}{2}\utilde{b}\;}\\ \;\\ \text{Now } 2\utilde{a}-3\utilde{b}=\utilde{0}\;\\ [/imath]
So if you allow the zero vector to be 'perpendicular' to a vector, the solution is [imath]\;\theta = 0 \text{ (and }\utilde{a}=\tfrac{3}{2}\utilde{b}\text{)}[/imath]
If you do not allow the zero vector to be 'perpendicular' to a vector, then there is no solution.

 

[Dr.Peterson]

You made a slight error here:

I'd rather say that the work so far is valid, but when you get to the end you appear to have a solution, [imath]\theta=0[/imath] -- except that it isn't.

This is in a category of problem that I find very tricky, where you can show that a solution is correct, assuming there is a solution at all, but the method you use is not in itself sufficient to show the existence of a solution.

I modeled this in GeoGebra, just to check how it works, and saw that the angle between 3a+2b and 2a-3b is always acute, approaching perpendicular as the latter approaches zero:

 

[lex]

I'd rather say that the work so far is valid, but when you get to the end you appear to have a solution, [imath]\theta=0[/imath] -- except that it isn't.

The problem does require some delicate argument.
Geogebra seems to be quite powerful - I must look into it.