Question

If $\vec u,\vec v,$and $\vec w$ are non-coplanar vectors and, $p$and $q$are real numbers, then the equality $\left[ {\begin{array}{*{20}{c}}
  {3\vec u}&{p\vec v}&{p\vec w}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {p\vec v}&{\vec w}&{q\vec u}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {2\vec w}&{q\vec v}&{q\vec u}
\end{array}} \right] = 0$ 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

Hint: The given equality $\left[ {\begin{array}{*{20}{c}}
  {3\vec u}&{p\vec v}&{p\vec w}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {p\vec v}&{\vec w}&{q\vec u}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {2\vec w}&{q\vec v}&{q\vec u}
\end{array}} \right] = 0$ is an equation consisting of scalar product. To solve this question we will use basic definition of scalar product i.e. \[\left[ {\begin{array}{*{20}{c}}
  {\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = \vec a.\left( {\vec b \times \vec c} \right)\] for any three vectors and properties of the scalar triple product.

Formula Used:
Three vectors are said to be non-coplanar if their support lines are not parallel to the same plane or they cannot be expressed as$\overrightarrow R = x\overrightarrow A + y\overrightarrow B + z\overrightarrow C $. Expanding the cross product of the expression, $(A + B + C) \cdot \left[ {A \times A + A \times C + B \times A + B \times C} \right]$by using identity $A \times A = 0$.

Complete step by step Solution:
In the question, the equality $\left[ {\begin{array}{*{20}{c}}
  {3\vec u}&{p\vec v}&{p\vec w}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {p\vec v}&{\vec w}&{q\vec u}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {2\vec w}&{q\vec v}&{q\vec u}
\end{array}} \right] = 0$_{is given,}
Before solving the given equation, we need to simplify the given equation using the properties of the scalar triple product. Few properties will be used here are listed below:
_1. $\left[ {\begin{array}{*{20}{c}}
  {\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\vec b}&{\vec c}&{\vec a}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\vec c}&{\vec a}&{\vec b}
\end{array}} \right]$
_2. $\left[ {\begin{array}{*{20}{c}}
  {\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = - \left[ {\begin{array}{*{20}{c}}
  {\vec b}&{\vec a}&{\vec c}
\end{array}} \right] = - \left[ {\begin{array}{*{20}{c}}
  {\vec c}&{\vec b}&{\vec a}
\end{array}} \right]$
3. Scalar multiplication:$\left[ {\begin{array}{*{20}{c}}
  {{k_1}\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = {k_1}\left[ {\begin{array}{*{20}{c}}
  {\vec a}&{\vec b}&{\vec c}
\end{array}} \right]$ where ${k_1}$is a scalar.
Consider the given equation, then we have:
$\left[ {\begin{array}{*{20}{c}}
  {3\vec u}&{p\vec v}&{p\vec w}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {p\vec v}&{\vec w}&{q\vec u}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {2\vec w}&{q\vec v}&{q\vec u}
\end{array}} \right] = 0$
Using property $(3)$, we can simplify the given equation in the form below;
$3{p^2}\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right] - pq\left[ {\begin{array}{*{20}{c}}
  {\vec v}&{\vec w}&{\vec u}
\end{array}} \right] - 2{q^2}\left[ {\begin{array}{*{20}{c}}
  {\vec w}&{\vec v}&{\vec u}
\end{array}} \right] = 0\,\,\,\,\,\,...(1)$
From property $(1)$, the second scalar triple product term can be replaced as follows: $\left[ {\begin{array}{*{20}{c}}
  {\vec v}&{\vec w}&{\vec u}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right]\,\,\,\,\,....(2)$
From property $(2)$, the third scalar triple product term can be replaced as follows:
$\left[ {\begin{array}{*{20}{c}}
  {\vec w}&{\vec v}&{\vec u}
\end{array}} \right] = - \left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right]\,\,\,\,\,\,....(3)$
Substituting the respective values from equation $(2)$ and $(3)$ in equation $(1)$, we get,
$3{p^2}\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right] - pq\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right] - 2{q^2}\left( { - \left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right]} \right) = 0$
Solving further, then:
$3{p^2}\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right] - pq\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right] + 2{q^2}\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right] = 0$
Taking $\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right]$ common,
$\left( {3{p^2} - pq + 2{q^2}} \right)\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right] = 0\,\,\,\,\,\,\,....(4)$
Now, since it is already given that the vectors $\vec u,\vec v,$and $\vec w$ are non-coplanar vectors, therefore$\left[ {\begin{array}{*{20}{c}}
  {\vec u}&{\vec v}&{\vec w}
\end{array}} \right] \ne 0$.
Then, from equation $(4)$ it implies that,
$\left( {3{p^2} - pq + 2{q^2}} \right) = 0\,\,\,\,\,\,...(5)$
We can find the relation between $p$and $q$by some algebraic manipulations to solve equation (5)
Using Completing the square method to solve for, we first divide the whole equation by $3$, we get,
${p^2} - p.\dfrac{q}{3} + \dfrac{2}{3}{q^2} = 0 \\$
 ${p^2} - 2.p.\dfrac{q}{6} + \dfrac{2}{3}{q^2} = 0 \\$
Now, adding and subtracting ${\left( {\dfrac{q}{6}} \right)^2}$, we have,
   ${p^2} - 2.p.\dfrac{q}{6} + {\left( {\dfrac{q}{6}} \right)^2} - {\left( {\dfrac{q}{6}} \right)^2} + \dfrac{2}{3}{q^2} = 0 \\$
  ${\left( {p - \dfrac{q}{6}} \right)^2} - \dfrac{{{q^2}}}{{36}} + \dfrac{2}{3}{q^2} = 0 \\$
Solving further we get,
${\left( {p - \dfrac{q}{6}} \right)^2} + \dfrac{{23}}{{36}}{q^2} = 0\,\,\,\,\,\,...(6)$
We observe that in the LHS of equation $(6)$, both terms are square terms which means they cannot be negative. Also, if none of the terms in LHS will give a negative term cancel another term. Since $p$ and $q$ are both real numbers, it would imply that equation $(6)$ can only hold when both $p$ and $q$ are equal to zero, i.e. $p = 0$and $q = 0$.
Therefore, there is exactly one value for $p$and $q$which will hold for equation $\left[ {\begin{array}{*{20}{c}}
  {3\vec u}&{p\vec v}&{p\vec w}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {p\vec v}&{\vec w}&{q\vec u}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {2\vec w}&{q\vec v}&{q\vec u}
\end{array}} \right] = 0$.

Hence, the correct option is (D).

Note:The given question can also be solved by manipulation the scalar triple product in the form \[\left[ {\begin{array}{*{20}{c}}
  {\vec a}&{\vec b}&{\vec c}
\end{array}} \right] = \vec a.\left( {\vec b \times \vec c} \right)\], but it would be a lengthy solution, making use of the properties of scalar triple products is much simpler to understand. Also, the given quadratic equation can be solved using quadratic formula but it will lengthen the solution.