If vec{u}, vec{v}, vec{w} are non-coplanar ve | vedclass

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(D) Given the equation: $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$Using the property of sca

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exactly one value of $(p, q)$

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(D) Given the equation: $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$Using the property of scalar triple product $[k\vec{a}, l\vec{b}, m\vec{c}] = klm[\vec{a}, \vec{b}, \vec{c}]$,we get:$3p^2[\vec{u}, \vec{v}, \vec{w}] - pq[\vec{v}, \vec{w}, \vec{u}] - 2q^2[\vec{w}, \vec{v}, \vec{u}] = 0$Since $[\vec{v}, \vec{w}, \vec{u}] = [\vec{u}, \vec{v}, \vec{w}]$ and $[\vec{w}, \vec{v}, \vec{u}] = -[\vec{u}, \vec{v}, \vec{w}]$,the equation becomes:$3p^2[\vec{u}, \vec{v}, \vec{w}] - pq[\vec{u}, \vec{v}, \vec{w}] + 2q^2[\vec{u}, \vec{v}, \vec{w}] = 0$$(3p^2 - pq + 2q^2)[\vec{u}, \vec{v}, \vec{w}] = 0$Since $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar,$[\vec{u}, \vec{v}, \vec{w}] 
eq 0$. Therefore:$3p^2 - pq + 2q^2 = 0$This is a quadratic equation in $p$. For $p$ to be a real number,the discriminant $D$ must be $\geq 0$:$D = (-q)^2 - 4(3)(2q^2) = q^2 - 24q^2 = -23q^2$For $D \geq 0$,we must have $-23q^2 \geq 0$,which implies $q^2 \leq 0$. Since $q$ is a real number,this is only possible if $q = 0$.Substituting $q = 0$ into $3p^2 - pq + 2q^2 = 0$,we get $3p^2 = 0$,so $p = 0$.Thus,the only solution is $(p, q) = (0, 0)$.

If $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $p, q$ are real numbers,then the equality $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$ holds for:

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