Vector question 2

daphnewoon · #1 $Old$ October 24th, 2009, 10:25 AM

Use cross product to find the sine of the angle between the vectors

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

earboth · #2 $Old$ October 24th, 2009, 11:22 AM

Quote:

Originally Posted by daphnewoon $View Post$

Use cross product to find the sine of the angle between the vectors

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

daphnewoon · #3 $Old$ October 24th, 2009, 08:55 PM

Hello, thanks for replying
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?

earboth · #4 $Old$ October 25th, 2009, 12:55 AM

Quote:

Originally Posted by daphnewoon $View Post$

Hello, thanks for replying
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 ...

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 $\sin(\vec u, \vec v)$