Question

If $\vec u,\vec v,\vec w$ are non-coplanar vectors and p, q are real numbers, then the equality $\left[ {3\vec u{\text{ }}p\vec v{\text{ }}p\vec w} \right] - \left[ {p\vec v{\text{ }}\vec w{\text{ }}q\vec u} \right] - \left[ {2\vec w{\text{ }}q\vec v{\text{ }}q\vec u} \right] = 0$ holds for,
$\left( a \right)$ Exactly one value of (p, q)
$\left( b \right)$ Exactly two values of (p, q)
$\left( c \right)$ More than two but not for all values of (p, q)
$\left( d \right)$ All values of (p, q)

Answer 1

Hint: In this particular type of question use the concept that when we rotate the vectors u, v and w in order than the dot product of vectors $\vec u.\vec v.\vec w = \vec v.\vec w.\vec u = \vec w.\vec u.\vec v$and if the order is different except these three orders than the dot product of these vectors is negative times of dot product of these vectors, so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
$\vec u,\vec v,\vec w$ are non-coplanar vectors, and p, q are real numbers.
Now we have to find out the condition on p and q such that $\left[ {3\vec u{\text{ }}p\vec v{\text{ }}p\vec w} \right] - \left[ {p\vec v{\text{ }}\vec w{\text{ }}q\vec u} \right] - \left[ {2\vec w{\text{ }}q\vec v{\text{ }}q\vec u} \right] = 0$ holds true.
The above equation is also written as,
$ \Rightarrow 3{p^2}\left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right] - pq\left[ {\vec v{\text{ }}\vec w{\text{ }}\vec u} \right] - 2{q^2}\left[ {\vec w{\text{ }}\vec v{\text{ }}\vec u} \right] = 0$
Now as we know that when we rotate the vectors u, v and w in order than the dot product of vectors $\vec u.\vec v.\vec w = \vec v.\vec w.\vec u = \vec w.\vec u.\vec v$and if the order is different except these three orders than the dot product of these vectors is negative times of dot product of these vectors.
$ \Rightarrow \vec u.\vec v.\vec w = \vec v.\vec w.\vec u = \vec w.\vec u.\vec v$, and $\vec w.\vec v.\vec u = - \vec u.\vec v.\vec w$ so use these values in the above equation we have,
\[ \Rightarrow 3{p^2}\left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right] - pq\left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right] - 2{q^2}\left[ { - \left( {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right)} \right] = 0\]
\[ \Rightarrow 3{p^2}\left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right] - pq\left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right] + 2{q^2}\left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right] = 0\]
\[ \Rightarrow \left( {3{p^2} - pq + 2{q^2}} \right)\left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right] = 0\]................... (1)
Now it is given that $\vec u,\vec v,\vec w$ are non-coplanar vectors, so \[\left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right]\] can never be zero.
$ \Rightarrow \left[ {\vec u{\text{ }}\vec v{\text{ }}\vec w} \right] \ne 0$
So if this is not equal to zero than from equation (1) we have,
\[ \Rightarrow 3{p^2} - pq + 2{q^2} = 0\]
Now divide by 3 throughout we have,
\[ \Rightarrow {p^2} - p\left( {\dfrac{q}{3}} \right) + \dfrac{{2{q^2}}}{3} = 0\]
Now make the complete square in p by adding and subtracting the square of the half of the coefficient of p so we have,
\[ \Rightarrow {p^2} - p\left( {\dfrac{q}{3}} \right) + \dfrac{{2{q^2}}}{3} + {\left( {\dfrac{q}{6}} \right)^2} - {\left( {\dfrac{q}{6}} \right)^2} = 0\]
\[ \Rightarrow {\left( {p - \dfrac{q}{6}} \right)^2} + \dfrac{{2{q^2}}}{3} - {\left( {\dfrac{q}{6}} \right)^2} = 0\]
\[ \Rightarrow {\left( {p - \dfrac{q}{6}} \right)^2} + \dfrac{{2{q^2}}}{3} - {\left( {\dfrac{q}{6}} \right)^2} = 0\]
\[ \Rightarrow {\left( {p - \dfrac{q}{6}} \right)^2} + \dfrac{{23{q^2}}}{{36}} = 0\]
Now as we see that in the above equation both of the terms are positive and square so for the solution of the equation each term separately equal to zero so we have,
$ \Rightarrow \dfrac{{23{q^2}}}{{36}} = 0$
$ \Rightarrow q = 0$
And ${\left( {p - \dfrac{q}{6}} \right)^2} = 0$
$ \Rightarrow {\left( {p - 0} \right)^2} = 0$
$ \Rightarrow p = 0$
So the solution of the equation (2) is (0, 0)
So for only one value of (p, q), $\left[ {3\vec u{\text{ }}p\vec v{\text{ }}p\vec w} \right] - \left[ {p\vec v{\text{ }}\vec w{\text{ }}q\vec u} \right] - \left[ {2\vec w{\text{ }}q\vec v{\text{ }}q\vec u} \right] = 0$ holds true.
Hence option (a) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that after simplifying the quadratic equation in such a way that both of the terms of the equation is positive and the equation is equal to zero, then both of the terms individually equates to zero in order to get the solution of the equation.