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

(B) exactly two values of ($p, q$)

(C) more than two but not all values of (p,q)

JEE Main · 2009 · 12 Apr 2009 · Offline · Vector Algebra

u, v, w

are non-coplanar vectors and

p, q

are real numbers, then the equality

[3 u p v p w] - [p v w q u] - [2 w q v q u] = 0

holds for

Answer: (A)

Expanding each scalar triple product and using that swapping two vectors changes its sign, the whole left side collapses to

(3 p^{2} - pq + 2 q^{2}) [u v w]

. Because the vectors are non-coplanar their box product is nonzero, so the condition becomes

3 p^{2} - pq + 2 q^{2} = 0

. Writing this as

2 p^{2} + (p - \frac{q}{2})^{2} + \frac{7 q ^{2}}{4} = 0

, a sum of squares, it can hold only when

p = 0

and

q = 0

, so the equality is satisfied for exactly one pair

(p, q)

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option