Question

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

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

Answer 1

Answer: (A)

Exactly one value of $(p,q)$

we have

$3p^2\left[\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right]-pq\left[\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right]+2q^2\left[\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right]=0$
$\Rightarrow 3p^2-pq+2q^2=0$
$p=\frac{q\pm \sqrt{1-24}q}{2\times 3}$
$p=\frac{q\pm \sqrt{23}iq}{2\times 3}$

$\Rightarrow p=0,q=0$
Hence, option '$A^{\prime }$ is correct.