Question

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

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

Answer 1

Answer: (D)

exactly one value of $(p,q)$

We have $\left[l\overline{a}m\overline{b}n\overline{c}\right]=lmn\left[\overline{a}\overline{b}\overline{c}\right]$ for scalars l, m, n
Also $\left[\overline{a}\overline{b}\overline{c}\right]=\left[\overline{b}\overline{c}\overline{a}\right]=\left[\overline{c}\overline{a}\overline{b}\right]$ (cyclic)
And $\left[\overline{a}\overline{b}\overline{c}\right]=-\left[\overline{a}\overline{c}\overline{b}\right]$
(Interchange of any two vectors)
$\left[3\overline{u}p\overline{v}p\overline{w}\right]-\left[p\overline{v}\overline{w}q\overline{w}\right]+2q^2\left[\overline{u}\overline{v}\overline{w}\right]=0$
$\Rightarrow 3p^2\left[\overline{u}\overline{v}\overline{w}\right]-pq\left[\overline{u}\overline{v}\overline{w}\right]+2q^2\left[\overline{u}\overline{v}\overline{w}\right]=0$
$\Rightarrow (3p^2-pq+2q^2)\left[\overline{u}\overline{v}\overline{w}\right]=0$ As $\overline{u}\overline{v}\overline{w}$ are non-coplanar,$\left[\overline{u}\overline{v}\overline{w}\right]\ne 0$ Hence $3p^2-pq+2q^2=0,p,q\in R$ As a quadratic in p, roots are real $\Rightarrow q^2-24q^2\ge 0\Rightarrow -23q^2\ge 0\Rightarrow q^2\le 0\Rightarrow q=0$ And thus $p=0$
Thus $(p,q)=(0,0)$ is the only possibility