justift whether ( u+v)dot(u cross v)=0 for all non-zero vectors u and v.

need help with this question, managed to finish rest in problem set. thanks for the help

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- Nov 2nd 2008, 12:42 PMmonkiusVectors
justift whether ( u+v)dot(u cross v)=0 for all non-zero vectors u and v.

need help with this question, managed to finish rest in problem set. thanks for the help - Nov 2nd 2008, 01:00 PMskeeter
let u = ai + bj + ck

v = di + ej + fk

u + v = (a+d)i + (b+e)j + (c+f)k

u X v = (bf - ce)i - (af - cd)j + (ae - bd)k

(u + v) dot (u X v) =

(a + d)(bf - ce) - (b + e)(af - cd) + (c + f)(ae - bd)

(abf + bdf - ace - cde) - (abf + aef - bcd - cde) + (ace + aef - bcd - bdf)

abf + bdf - ace - cde - abf - aef + bcd + cde + ace + aef - bcd - bdf

believe all these cancel out to 0. - Nov 2nd 2008, 01:05 PMPlato
$\displaystyle \left( {A + B} \right) \cdot C = A \cdot C + B \cdot C$

$\displaystyle A \cdot \left( {A \times B} \right) = 0\,\& \,B \cdot \left( {A \times B} \right)= 0$

Applying these two will answer your question. - Nov 3rd 2008, 03:55 AMHallsofIvy
It also helps to know that $\displaystyle <x_1,y_1,z_1>\cdot(<x_2,y_2,z_2>\times<x_3,y_3,z_3 >)$, the "triple product", can be written in the determinant form:

$\displaystyle \left|\begin{array}{ccc}x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3\end{array}\right|$

In the case of either $\displaystyle u\cdot(u\times v)$ or $\displaystyle v\cdot(u\times v)$, two of the rows of the determinant are the same and so the determinant is 0.