Question
If $$\vec u,\,\vec v,\,\vec \omega $$ are non-coplanar vectors and $$p,\,q$$ are real numbers, then the equality $$\left[ {3\vec u\,p\vec v\,p\vec \omega } \right] - \left[ {p\vec v\,\vec \omega \,q\vec u} \right] - \left[ {2\vec \omega \,q\vec v\,q\vec u} \right] = 0$$ holds for :
A.
exactly two values of $$\left( {p,\,q} \right)$$
B.
more than two but not all values of $$\left( {p,\,q} \right)$$
C.
all values of $$\left( {p,\,q} \right)$$
D.
exactly one value of $$\left( {p,\,q} \right)$$
Answer :
exactly one value of $$\left( {p,\,q} \right)$$
Solution :
$$\eqalign{
& \left[ {3\vec u\,p\vec v\,p\vec \omega } \right] - \left[ {p\vec v\,\vec \omega \,q\vec u} \right] - \left[ {2\vec \omega \,q\vec v\,q\vec u} \right] = 0 \cr
& \Rightarrow \left( {3{p^2} - pq + 2{q^2}} \right)\left[ {\vec u\,\vec v\,\vec \omega } \right] = 0 \cr
& \Rightarrow 3{p^2} - pq + 2{q^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\left[ {\vec u\,\vec v\,\vec \omega } \right] \ne 0} \right] \cr
& \Rightarrow 2{p^2} + {p^2} - pq + \frac{{{q^2}}}{4} + \frac{{7{q^2}}}{4} = 0 \cr
& \Rightarrow 2{p^2} + {\left( {p - \frac{q}{2}} \right)^2} + \frac{7}{4}{q^2} = 0 \cr
& \Rightarrow p = 0,\,\,q = 0,\,\,p = \frac{q}{2} \cr
& \Rightarrow p = 0,\,\,q = 0 \cr} $$
$$\therefore $$ Exactly one value of $$\left( {p,\,q} \right)$$