Scalar triple product coplanarity

Neen87
Messages
8
Reaction score
0

Homework Statement



Show that u, v, w lie in the same plane in R3 if and only if u · (v × w) = 0.


Homework Equations





The Attempt at a Solution



if u · (v × w) = 0, then u is orthogonal to vxw, and
vxw is orthogonal to v and w.

therefore, u must lie in the same plane determined by v and w (i.e. they are coplanar)


Is this correct? Also, how would i describe this proof mathematically?

Thanks!
 
Physics news on Phys.org
u · (v × w) = 0

use geometrical proof

translate the equation as volume of parallopiped
the volume is zero only when the height is zero
or all vectors lie in the same plane
 
Neen87 said:

Homework Statement



Show that u, v, w lie in the same plane in R3 if and only if u · (v × w) = 0.


Homework Equations





The Attempt at a Solution



if u · (v × w) = 0, then u is orthogonal to vxw, and
vxw is orthogonal to v and w.

therefore, u must lie in the same plane determined by v and w (i.e. they are coplanar)


Is this correct? Also, how would i describe this proof mathematically?

Thanks!
Yes, that's correct and, once you have stated exactly HOW you know that two vectors, perpendicular to the same non-zero vector, are co-planar, it IS "mathematical".

payumooli's suggestion is another way to do it but since you have already done it your own way, I suggest you stay with it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top