help please!

[hayood]

Let u, v, and w be nonzero vectors in 3-space with the same initial point, but such that no two of them are collinear. Show that
(a) u x (v x w) lies in the plane determined by v and w
(b) (u x v) x w lies in the plane determined by u and v

 

[soroban]

Hello, hayood!

$\displaystyle \text{Let }\vec u,\,\vec v,\,\vec w \text{ be nonzero vectors in 3-space with the same initial point,}$
. . $\displaystyle \text{such that no two of them are collinear.}$

$\displaystyle \text{Show that:}$

$\displaystyle \text{(a) }\:\vec u \times (\vec v \times \vec w)\,\text{ lies in the plane determined by }\vec v\text{ and }\vec w.$

\(\displaystyle \text{(b) }\\vec u \times \vec v) \times \vec w\text{ lies in the plane determined by }\vec u\text{ and }\vec v.\)

Can it be this easy? . . .

          |
          |
    v x w |        * . . . . .
          |   w *         . 
          |  *          . 
          o  *  *  *  *
                 v

$\displaystyle (\vec v \times \vec w )\,\text{ is normal to the plane of }\vec v\text{ and }\vec w.$

$\displaystyle \text{And: }\:\vec u \times (\vec v \times \vec w)\,\text{ is normal to the plane of }\vec u\text{ and }(\vec v \times \vec w).$

$\displaystyle \text{Hence: }\:\vec u \times (\vec v \times \vec w)\,\text{ is }parallel\text{ to the plane of }\vec v\text{ and }\vec w.$


$\displaystyle \text{Since all the vectors have the same inital point,}$

. . $\displaystyle \vec u \times (\vec v \times \vec w)\,\text{ lies }in\text{ the plane of }\vec v\text{ and }\vec w.$