How to Find the Angle Between Two Vectors?

[Michaelh926]

Question:
Let u·v = 5 and u×v = 2i+3j+k. Determine the angle (in degrees to one
decimal place) between the two vectors.

Im not sure on what to do with the equation given since other examples that were done i was given separate equations for u and v

We were told this formula will be used but not sure how to get u and v.
|u x v|=|u||v|sinθ

Thanks in advance to anyone that helps

 

[Andrew Mason]

Question:
Let u·v = 5 and u×v = 2i+3j+k. Determine the angle (in degrees to one
decimal place) between the two vectors.

Im not sure on what to do with the equation given since other examples that were done i was given separate equations for u and v

We were told this formula will be used but not sure how to get u and v.
|u x v|=|u||v|sinθ

Thanks in advance to anyone that helps

What is the value of |u x v|? From that you should be able to work out the value of tanθ and then θ without needing to find u or v.

AM

 

[Michaelh926]

What is the value of |u x v|?

so, the value of |u x v|=sqrt(2^+3^+1^)?

then substitute it so it will be |sqrt(2^+3^+1^)|=|u||v|sinθ?

 

[Michaelh926]

What is the value of |u x v|? From that you should be able to work out the value of tanθ and then θ without needing to find u or v.

AM

so, the value of |u x v|=sqrt(2^+3^+1^)?

then substitute it so it will be |sqrt(2^+3^+1^)|=|u||v|sinθ?

 

[Curious3141]

so, the value of |u x v|=sqrt(2^+3^+1^)?

then substitute it so it will be |sqrt(2^+3^+1^)|=|u||v|sinθ?

You're missing the exponents of 2 (squares), but yes.

Now find a similar expression for the dot product, divide and form a simple trig equation.

 

[Michaelh926]

You're missing the exponents of 2 (squares), but yes.

Now find a similar expression for the dot product, divide and form a simple trig equation.

Sorry, forgot to put the exponents in.

So will this be what it comes to...
sinθ=sqrt(14)/5 then solve for theta?

 

[Andrew Mason]

Sorry, forgot to put the exponents in.

So will this be what it comes to...
sinθ=sqrt(14)/5 then solve for theta?

Not quite. You said:

(1) uvsinθ = √14

(2) uvcosθ = 5


Solve for θ.

AM

 

[Curious3141]

Sorry, forgot to put the exponents in.

So will this be what it comes to...
sinθ=sqrt(14)/5 then solve for theta?

No. Isn't there a cos θ in the definition of the dot product?

Andrew Mason's post gives you the equations you should divide.

 

[Michaelh926]

No. Isn't there a cos θ in the definition of the dot product?

Andrew Mason's post gives you the equations you should divide.

so is it

√14/sinθ=5/cosθ
tanθ=√14/5
θ=arctan(√14/5)?

 

[Curious3141]

so is it

√14/sinθ=5/cosθ
tanθ=√14/5
θ=arctan(√14/5)?

Yes.

 

[Michaelh926]

Yes.

:) now i understand it.

thank you to everyone that helped