Question

If $\overline{u},\overline{v},\overline{w}$ are non-coplanar vectors and p,q are real numbers, then the equality
$\left[3\overline{u}p\overline{v}p\overline{w}\right]-\left[p\overline{v}\overline{w}q\overline{u}\right]-\left[2\overline{w}q\overline{v}q\overline{u}\right]=0$ hold for

• A. Exactly one value of $(p,q)$
• B. Exactly two value of $(p,q)$
• C. More than two but not all values of $(p,q)$
• D. All value of $(p,q)$

Answer 1

The correct option is A Exactly one value of $(p,q)$

$\left[3\overline{u}p\overline{v}p\overline{w}\right]-\left[p\overline{v}\overline{w}q\overline{u}\right]-\left[2\overline{w}q\overline{v}q\overline{u}\right]=0$

$\Rightarrow 3p^2\left[\overline{u}\overline{v}\overline{w}\right]-pq\left[\overline{u}\overline{v}\overline{w}\right]+2q^2\left[\overline{u}\overline{v}\overline{w}\right]=0$

$\Rightarrow (3p^2-pq+2q^2)\left[\overline{u}\overline{v}\overline{w}\right]=0$

$\left[\overline{u}\overline{v}\overline{w}\right]\ne 0$

$\therefore 3p^2-pq+2q^2=0$

$\Rightarrow q^2-24q^2\ge 0\Rightarrow -23q^2\ge 0$

$q^2\le 0\Rightarrow q=0$

And thus $p=0$

Thus $(p,q)\equiv (0,0)$ is only the possibility.