1. Quick vector question

Hi
I need help on the following question:

Find a unit vector perpendicular to the plane of the two vectors (2,-1,1) and (3,4,-1). What is the sine of the angle between these two vectors?

I have found that the unit vector perpendicular to the plane is:
$\displaystyle \frac{-3i+5j+11k}{\sqrt{155}}$

However how would i work out the angle. i used the $\displaystyle a x b = absin\theta$ equation but cannot get the right answer.

P.S

2. Originally Posted by Paymemoney
Hi
I need help on the following question:

Find a unit vector perpendicular to the plane of the two vectors (2,-1,1) and (3,4,-1). What is the sine of the angle between these two vectors?

I have found that the unit vector perpendicular to the plane is:
$\displaystyle \frac{-3i+5j+11k}{\sqrt{155}}$

However how would i work out the angle. i used the $\displaystyle a x b = absin\theta$ equation but cannot get the right answer.

P.S
Use the definition of the scalar product:

$\displaystyle \vec a \cdot \vec b = |\vec a| \cdot |\vec b| \cdot \cos(\angle(\vec a, \vec b))$

Solve for $\displaystyle \cos(\angle(\vec a, \vec b))$ and afterwards determine $\displaystyle \angle(\vec a, \vec b)$

3. Originally Posted by earboth
Use the definition of the scalar product:

$\displaystyle \vec a \cdot \vec b = |\vec a| \cdot |\vec b| \cdot \cos(\angle(\vec a, \vec b))$

Solve for $\displaystyle \cos(\angle(\vec a, \vec b))$ and afterwards determine $\displaystyle \angle(\vec a, \vec b)$
why would i use the scalar form? It is asking to find the sine angle.

4. Why not? You do know that $\displaystyle sin x= \sqrt{1- cos^2 x}$, don't you?

Either way will give the correct answer. Since you don't show what you did or what answer you got, no one can tell what you did wrong.

5. ok i used the formula which i said and i get the answer:
a x b = $\displaystyle (absin\theta)$ * unit vector of a,b

$\displaystyle \frac{a * b}{unit vector of a,b} = absin\theta$

$\displaystyle \frac{-3i+5j+11k}{1} * \frac{\sqrt{155}}{-3i+5j+11k} = absin\theta$

-3i+5j+11k cancels so i am left with:

$\displaystyle \frac{\sqrt{155}}{\sqrt{156}} = sin\theta$

therefore $\displaystyle sin\theta = 0.9968$

6. so i used the way you said and i get the same answer:

$\displaystyle a \cdot b = abcos\theta$

$\displaystyle cos\theta = \frac{1}{\sqrt{156}}$

sub into $\displaystyle sin\theta = \sqrt{1-cos^{2}\theta}$

so $\displaystyle sin\theta = \sqrt{1-0.08006^2}$

$\displaystyle sin\theta = 0.9968$