Vector Algebra Question 87
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)
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ \because \vec{u},,\vec{v},,\vec{w} $ are non-coplanar vectors $ \because [ \vec{u},,\vec{v},,\vec{w} ]\ne 0 $ Now, $ [ 3\vec{u},,p\vec{v},,p\vec{w} ]-[ p\vec{v},p\vec{w},q\vec{u} ]-[ 2\vec{w},q\vec{v},q\vec{u} ]=0 $
$ \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
