Question

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

The Correct Option is A

$\left(3p^{2}-pq+2q^{2}\right)\left[\overrightarrow{u}\,\overrightarrow{v}\,\overrightarrow{w}\,\right]=0$ But $\left[\overrightarrow{u}\,\overrightarrow{v}\,\overrightarrow{w}\,\right]\ne0$ $3p^{2}-pq+2q^{2}=0$ $2p^{2}+p^{2}-pq+\left(\frac{q}{2}\right)^{2}+\frac{7q^{2}}{4}=0 \Rightarrow 2p^{2}+\left(p-\frac{q}{2}\right)^{2}+\frac{7}{4}q^{2}=0$ $\Rightarrow p = 0, q = 0, p=\frac{q}{2}$ This possible only when $p = 0, q = 0$ exactly one value of $\left(p, q\right)$