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.
https://www.physicsforums.com/threads/cross-product-homework.352545/



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

 

[Charles Link]

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 $V=T \times U$.

 

[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?

 

[Student100]

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.

 

[Student100]

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?

 

[Jovy]

Yes, I just forgot to take the arcsin of that. Thank you!