Question

If \( \vec{u}, \vec{v}, \vec{w} \) are non-coplanar vectors and \( p, q \) are real numbers, then the equality:

\[ [3\vec{u} \quad p\vec{v} \quad p\vec{w}] - [p\vec{v} \quad \vec{w} \quad q\vec{u}] - [2\vec{w} \quad q\vec{v} \quad q\vec{u}] = 0 \]

holds for:

• A. exactly one ordered pair of \( (p, q) \)
• B. exactly two ordered pairs of \( (p, q) \)
• C. all ordered pairs of \( (p, q) \)
• D. no ordered pair of \( (p, q) \)

Hint:

When dealing with scalar triple products, use linearity and the antisymmetric property: \[ [\vec{a}
\vec{b}
\vec{c}] = -[\vec{b}
\vec{a}
\vec{c}] \] For solving homogenous equations in multiple variables, substitution with ratios (e.g., \( p = kq \)) helps reduce complexity.

Answer 1

The Correct Option is A

Step 1: Use scalar triple product identity
The scalar triple product is defined as: \[ [\vec{a}
\vec{b}
\vec{c}] = \vec{a} . (\vec{b} \times \vec{c}) \] Given that \( \vec{u}, \vec{v}, \vec{w} \) are non-coplanar, their scalar triple product is non-zero. Use linearity of scalar triple product: \[ [3\vec{u}
p\vec{v}
p\vec{w}] = 3p^2[\vec{u}
\vec{v}
\vec{w}] \] \[ [p\vec{v}
\vec{w}
q\vec{u}] = pq[\vec{v}
\vec{w}
\vec{u}] = -pq[\vec{u}
\vec{v}
\vec{w}] \] \[ [2\vec{w}
q\vec{v}
q\vec{u}] = 2q^2[\vec{w}
\vec{v}
\vec{u}] = -2q^2[\vec{u}
\vec{v}
\vec{w}] \] So the full expression becomes: \[ 3p^2[\vec{u}
\vec{v}
\vec{w}] - (-pq[\vec{u}
\vec{v}
\vec{w}]) - (-2q^2[\vec{u}
\vec{v}
\vec{w}]) \] \[ = (3p^2 + pq + 2q^2)[\vec{u}
\vec{v}
\vec{w}] \] Set this equal to 0: \[ (3p^2 + pq + 2q^2)[\vec{u}
\vec{v}
\vec{w}] = 0 \Rightarrow 3p^2 + pq + 2q^2 = 0 \] Step 2: Solve the quadratic in two variables
We solve for real values \( (p, q) \) that satisfy: \[ 3p^2 + pq + 2q^2 = 0 \] This is a homogeneous quadratic in \( p \) and \( q \). Try \( p = kq \) and substitute: \[ 3k^2q^2 + kq^2 + 2q^2 = 0 \Rightarrow q^2(3k^2 + k + 2) = 0 \Rightarrow 3k^2 + k + 2 = 0 \] Solve using quadratic formula: \[ k = \frac{-1 \pm \sqrt{1^2 - 4(3)(2)}}{2(3)} = \frac{-1 \pm \sqrt{1 - 24}}{6} = \frac{-1 \pm \sqrt{-23}}{6} \] No real solution the equation has only one real solution when we solve directly in terms of \( (p, q) \). Try solving: \[ 3p^2 + pq + 2q^2 = 0 \] Using discriminant method (treat as quadratic in \( p \)): \[ \Delta = (q)^2 - 4(3)(2q^2) = q^2 - 24q^2 = -23q^2<0 \] No real solutions unless \( q = 0 \). Try \( q = 0 \Rightarrow 3p^2 = 0 \Rightarrow p = 0 \) So only solution: \( (p, q) = (0, 0) \) Exactly one ordered pair of \( (p, q) \) satisfies the equation.