Key Concepts and Formulas

• Scalar Triple Product (STP): The scalar triple product of three vectors $\overrightarrow a, \overrightarrow b, \overrightarrow c$ is denoted by $[\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c ]$ and represents the volume of the parallelepiped formed by these vectors.
• Properties of STP:
  • Scalar Multiplication: $[k_1 \overrightarrow a \,\, k_2 \overrightarrow b \,\, k_3 \overrightarrow c ] = k_1 k_2 k_3 [\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c ]$.
  • Cyclic Permutation: $[\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c ] = [\overrightarrow b \,\,\overrightarrow c \,\,\overrightarrow a ] = [\overrightarrow c \,\,\overrightarrow a \,\,\overrightarrow b ]$.
  • Non-coplanarity: If $\overrightarrow u, \overrightarrow v, \overrightarrow w$ are non-coplanar, then $[\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] \neq 0$.

Step-by-Step Solution

We are given the equation: $\left[ {3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow w } \right] - \left[ {p\overrightarrow v \,\,\overrightarrow w \,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow w \,\,q\overrightarrow v \,\,q\overrightarrow u } \right] = 0$ We are also given that $\overrightarrow u, \overrightarrow v, \overrightarrow w$ are non-coplanar vectors, which implies that $[\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] \neq 0$.

Step 1: Apply the scalar multiplication property of the scalar triple product to each term in the given equation. The first term is $[3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow w ]$. Using the property, we can take out the scalars 3, p, and p: $[3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow w ] = 3 \cdot p \cdot p [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] = 3p^2 [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ]$ The second term is $[p\overrightarrow v \,\,\overrightarrow w \,\,q\overrightarrow u ]$. Taking out the scalars p, 1, and q: $[p\overrightarrow v \,\,\overrightarrow w \,\,q\overrightarrow u ] = p \cdot 1 \cdot q [\overrightarrow v \,\,\overrightarrow w \,\,\overrightarrow u ] = pq [\overrightarrow v \,\,\overrightarrow w \,\,\overrightarrow u ]$ The third term is $[2\overrightarrow w \,\,q\overrightarrow v \,\,q\overrightarrow u ]$. Taking out the scalars 2, q, and q: $[2\overrightarrow w \,\,q\overrightarrow v \,\,q\overrightarrow u ] = 2 \cdot q \cdot q [\overrightarrow w \,\,\overrightarrow v \,\,\overrightarrow u ] = 2q^2 [\overrightarrow w \,\,\overrightarrow v \,\,\overrightarrow u ]$ So the equation becomes: $3p^2 [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] - pq [\overrightarrow v \,\,\overrightarrow w \,\,\overrightarrow u ] - 2q^2 [\overrightarrow w \,\,\overrightarrow v \,\,\overrightarrow u ] = 0$

Step 2: Use the cyclic permutation property of the scalar triple product to express all terms with the same base STP, i.e., $[\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ]$. We know that $[\overrightarrow v \,\,\overrightarrow w \,\,\overrightarrow u ] = [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ]$ (one cyclic shift). We also know that $[\overrightarrow w \,\,\overrightarrow v \,\,\overrightarrow u ] = -[\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ]$ (swapping two vectors negates the STP). Substituting these into the equation from Step 1: $3p^2 [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] - pq [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] - 2q^2 (-[\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ]) = 0$

Step 3: Simplify the equation by factoring out the common term $[\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ]$. $3p^2 [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] - pq [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] + 2q^2 [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] = 0$ $(3p^2 - pq + 2q^2) [\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] = 0$

Step 4: Since $\overrightarrow u, \overrightarrow v, \overrightarrow w$ are non-coplanar, we have $[\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ] \neq 0$. Therefore, for the equation to hold, the scalar factor must be zero. $3p^2 - pq + 2q^2 = 0$

Step 5: Analyze the quadratic form $3p^2 - pq + 2q^2 = 0$ to determine the values of $p$ and $q$ that satisfy it. We can treat this as a quadratic equation in $p$ (or $q$). Let's consider it as a quadratic in $p$: $3p^2 - (q)p + 2q^2 = 0$. For real solutions of $p$, the discriminant of this quadratic must be non-negative. The discriminant is given by $\Delta = b^2 - 4ac$, where $a=3$, $b=-q$, and $c=2q^2$. $\Delta = (-q)^2 - 4(3)(2q^2)$ $\Delta = q^2 - 24q^2$ $\Delta = -23q^2$ For real values of $p$, we need $\Delta \ge 0$. $-23q^2 \ge 0$ Since $q^2 \ge 0$ for any real number $q$, the term $-23q^2$ is always less than or equal to 0. The only way for $-23q^2 \ge 0$ to be true is if $q^2 = 0$, which implies $q=0$.

If $q=0$, substitute this back into the equation $3p^2 - pq + 2q^2 = 0$: $3p^2 - p(0) + 2(0)^2 = 0$ $3p^2 = 0$ This implies $p=0$.

So, the only real solution for $(p,q)$ that satisfies $3p^2 - pq + 2q^2 = 0$ is $(p,q) = (0,0)$.

Step 6: Interpret the result in terms of the number of possible values for $(p,q)$. We found that the equality holds only for the specific pair of values $(p,q) = (0,0)$. This means there is exactly one pair of values $(p,q)$ for which the given equality holds.

Common Mistakes & Tips

• Incorrect application of STP properties: Ensure that scalar multiplication and cyclic permutation properties are applied correctly, especially with regard to signs when vectors are swapped.
• Ignoring the non-coplanarity condition: The condition that vectors are non-coplanar is crucial, as it allows us to divide by $[\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow w ]$ (or more precisely, to set the scalar coefficient to zero).
• Mistakes in analyzing the quadratic form: When analyzing $3p^2 - pq + 2q^2 = 0$, it's important to correctly calculate the discriminant and understand its implications for real solutions. A common mistake is to assume that if the discriminant is negative, there are no solutions, without considering the case where the expression is identically zero. In this case, if $q \neq 0$, the discriminant is strictly negative, meaning no real solutions for $p$. If $q=0$, the discriminant is zero, and we get a solution.

Summary

The problem involves simplifying a given equation involving scalar triple products of non-coplanar vectors. By applying the properties of scalar triple products, the equation reduces to a condition on the real numbers $p$ and $q$: $3p^2 - pq + 2q^2 = 0$. Analyzing this quadratic form reveals that the only real solution for $(p,q)$ is $(0,0)$. Therefore, the equality holds for exactly one pair of values of $(p,q)$.

The final answer is \boxed{A}.