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

hold for

 

  • Option 1)

    exactly two values of  \left ( p,q \right )

  • Option 2)

    more than two but not all values of \left ( p,q \right )

  • Option 3)

    all values of  \left ( p,q \right )

  • Option 4)

    exactly one value of  \left ( p,q \right )

 

Answers (1)

As we learnt in 

Scalar Triple Product -

\left [ \vec{a}\;\vec{b}\; \vec{c} \right ]

=\left (\vec{a}\times \vec{b}\right)\cdot \vec{c}= \vec{a}\cdot \left ( \vec{b} \times \vec{c}\right )

=\left (\vec{b}\times \vec{c}\right)\cdot \vec{a}= \vec{b}\cdot \left ( \vec{c} \times \vec{a}\right )

=\left (\vec{c}\times \vec{a}\right)\cdot \vec{b}= \vec{c}\cdot \left ( \vec{a} \times \vec{b}\right )

- wherein

Scalar Triple Product of three vectors \hat{a},\hat{b},\hat{c}.

 

 (3p^{2}-pq+2q^{2})\:[\vec{u}\:\vec{v}\:\vec{w}]\:=0

But [\vec{u}\:\vec{v}\:\vec{w}]\neq 0

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

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

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

p=0,\ q=0,\ p=\frac{q}{2}


Option 1)

exactly two values of  \left ( p,q \right )

This option is incorrect.

Option 2)

more than two but not all values of \left ( p,q \right )

This option is incorrect.

Option 3)

all values of  \left ( p,q \right )

This option is incorrect.

Option 4)

exactly one value of  \left ( p,q \right )

This option is correct.

Posted by

Vakul

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