# Vectors: Cross Product

Jovy
Hello, I apologize in advance for the way this post looks. I am new to this forum and I've never used LaTeX Primer. I noticed that someone has prevoiusly asked the same question, but I still do not understand how to get to the answer. Also, I tried posting an image but I could not; and this other post contains the image so I will leave a link to the other forum.

1. Homework Statement

The figure shows two vectors $$\vec T$$ and $$\vec U$$ separated by an angle $$θ_{TU}$$ You are given that $$\vec T =(3,1,0), \vec U =(2,4,0), ~and~ \vec T × \vec U = \vec V$$ Find the sine of the angle between $$\vec T ~and~ \vec U$$

## Homework Equations

$$| \vec T × \vec U |= | \vec T || \vec V |\sinθ$$

## The Attempt at a Solution

I know that the answer is $$\frac {\sqrt{2}} 2$$ The magnitude of vector V is 10 and you square root that, 10 is also opposite of the angle theta. However, I don't understand how the hypotenuse is 20 therefore you would simplify $$\frac {\sqrt{10}} {\sqrt20}$$ to get the answer

Homework Helper
Gold Member
You didn't compute ## T \times U ## as far as I can see. This is a necessary computation. The easiest way that I know is by the determinant method. Have they shown you the determinant method in your coursework? ## \\ ## Editing... Looking over your work, the correct equation is ## | T \times U|=|T||U| sin(\theta) ##. (The equation you have written is incorrect.) The correct result is thereby ## \frac{|V|}{|T||U|}=sin(\theta) ## where ## V=T \times U ##.

Last edited:
Jovy
Jovy
You didn't compute ## T \times U ## as far as I can see. This is a necessary computation. The easiest way that I know is by the determinant method. Have they shown you the determinant method in your coursework? ## \\ ## Editing... Looking over your work, the correct equation is ## | T \times U|=|T||U| sin(\theta) ##. (The equation you have written is incorrect.) The correct result is thereby ## \frac{|V|}{|T||U|}=sin(\theta) ## where ## V=T \times U ##.

Isn't the magnitude of T ##\sqrt{3^2+1^2+0^2}## = ##\sqrt{10}## and the magnitude of U ##\sqrt{2^2+4^2+0^2}## = ##\sqrt{20}## ? But if you plug that into ## \frac{|V|}{|T||U|}=sin(\theta) ## , the answer is not correct. Where am I going wrong?

Gold Member
Isn't the magnitude of T ##\sqrt{3^2+1^2+0^2}## = ##\sqrt{10}## and the magnitude of U ##\sqrt{2^2+4^2+0^2}## = ##\sqrt{20}## ? But if you plug that into ## \frac{|V|}{|T||U|}=sin(\theta) ## , the answer is not correct. Where am I going wrong?

It gives you the correct answer.

What is ##\vec{T} \times \vec{U}##?

Jovy
It gives you the correct answer.

What is ##\vec{T} \times \vec{U}##?

It is (0,0,10), so the magnitude of that is 10.

Gold Member
It is (0,0,10), so the magnitude of that is 10.

Yes, sooo...

$$\frac{|V|}{|T||U|} = \sin(\theta)$$
$$\arcsin(\frac{|V|}{|T||U|}) = \theta$$
$$\arcsin(\frac{\sqrt{100}}{\sqrt{10}\sqrt{20}}) = \theta$$
$$\arcsin(\sqrt{\frac{100}{200}}) = \theta$$

Can you finish it from there?