Vector-Algebra Question 396
Question: 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{\omega },q\vec{u}]-[2\vec{\omega },q\vec{v},q\vec{u}]=0 $ holds for:
Options:
A) Exactly two values of (p, q)
B) More than two but not all values of (p, q)
C) All values of (p, q)
D) Exactly one value of (p, q)
Correct Answer: DShow Answer
Answer:
Solution:
$ \Rightarrow 3p^{2}[ \vec{u},\vec{v},\vec{w} ]-pq[ \vec{v},\vec{w},\vec{u} ]-2q^{2}[ \vec{w},\vec{v},\vec{u} ]=0 $
$ \Rightarrow 3p^{2}[ \vec{u},\vec{v},\vec{w} ]-pq[ \vec{u},\vec{v},\vec{w} ]-2q^{2}[ \vec{u},\vec{v},\vec{w} ] $
$ \Rightarrow (3p^{2}-pq+2q^{2})[ \vec{u},\vec{v},\vec{w} ]=0 $
$ \Rightarrow 3p^{2}-pq+2q^{2}=0 $
$ \Rightarrow 2p^{2}+p^{2}-pq+\frac{q^{2}}{4}+\frac{7q^{2}}{4}=0 $
$ \Rightarrow 2p^{2}+{{( p-\frac{q}{2} )}^{2}}+\frac{7}{4}q^{2}=0\Rightarrow p=0,q=0, $ $ p=q/2 $ This is possible only when $ p=0,q=0 $
$ \therefore $ There is exactly one value of (p, q).
BETA
