# Vectors

• Nov 2nd 2008, 01:42 PM
monkius
Vectors
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, 02:00 PM
skeeter
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, 02:05 PM
Plato
$\left( {A + B} \right) \cdot C = A \cdot C + B \cdot C$

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

Applying these two will answer your question.
• Nov 3rd 2008, 04:55 AM
HallsofIvy
It also helps to know that $\cdot(\times)$, the "triple product", can be written in the determinant form:
$\left|\begin{array}{ccc}x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3\end{array}\right|$

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