# Vector question 2

• Oct 24th 2009, 09:25 AM
daphnewoon
Vector question 2
Use cross product to find the sine of the angle between the vectors

u (2,3,-6)
v(2,3,6)
• Oct 24th 2009, 10:22 AM
earboth
Quote:

Originally Posted by daphnewoon
Use cross product to find the sine of the angle between the vectors

u (2,3,-6)
v(2,3,6)

Use the definition of the cross product:

$|\vec u \times \vec v|= |\vec u| \cdot |\vec v| \cdot \sin(\vec u, \vec v)$

Solve for $\sin(\vec u, \vec v)$
• Oct 24th 2009, 07:55 PM
daphnewoon
I know the formula to solve this question, just that i've gt this equation:

(36 , -30, 0) = 10 sin alpha

How am i going to manipulate this?
• Oct 24th 2009, 11:55 PM
earboth
Quote:

Originally Posted by daphnewoon
I know the formula to solve this question, just that i've gt this equation:

(36 , -30, 0) = 10 sin alpha

How am i going to manipulate this?

I'm not sure how you got this result ... (Thinking)

Use the formula I posted in my previous answer:

$
|\vec u \times \vec v|= |\vec u| \cdot |\vec v| \cdot \sin(\vec u, \vec v)
$

Plug in all the values you know:

$|(2,3,-6) \times (2,3,6)|= |(2,3,-6)| \cdot |(2,3,6)| \cdot \sin(\vec u, \vec v)$

$|(36, -24, 0)|= 7 \cdot 7 \cdot \sin(\vec u, \vec v)$

$12 \cdot \sqrt{13}= 49 \cdot \sin(\vec u, \vec v)$

... and now solve for $\sin(\vec u, \vec v)$