If u, v, w are non-coplanar vectors and p,q are real numbers, then the equality [3 u p v p…

Question

If $\vec u,\vec v,\vec w$ are non-coplanar vectors and $p,q$ are real numbers, then the equality $[3\vec u\ \ p\vec v\ \ p\vec w]-[p\vec v\ \ \vec w\ \ q\vec u]-[2\vec w\ \ q\vec v\ \ q\vec u]=0$ holds for:

Accepted Answer

$$3p^2[\vec u\ \vec v\ \vec w]-pq[\vec v\ \vec w\ \vec u]-2q^2[\vec w\ \vec v\ \vec u]=0$$

Using $[\vec v\ \vec w\ \vec u]=[\vec u\ \vec v\ \vec w]$ and $[\vec w\ \vec v\ \vec u]=-[\vec u\ \vec v\ \vec w]$:

$$[\vec u\ \vec v\ \vec w](3p^2-pq+2q^2)=0$$

Since $\vec u,\vec v,\vec w$ are non-coplanar, $3p^2-pq+2q^2=0$. Treating as a quadratic in $p$, the discriminant is $q^2-24q^2=-23q^2\le 0$, equality only at $q=0$, which forces $p=0$. So exactly one value, $(p,q)=(0,0)$.

Solutions · 1