Question

Let $u$ and $v$ be unit vectors. If $w$ is a vector such that $w+(w\times u)=v$, then $|(u\times v)\cdot w|$

• A. $\le \frac{1}{3}$
• B. $\le \frac{1}{2}$
• C. $>\frac{1}{3}$
• D. $\ge \frac{1}{2}$

Answer 1

Answer: (B)

Given equation is $w+(w\times u)=v(1)$
Taking cross-product with $u$, we get
$u\times [w+(w\times u)]=u\times v$
$\Rightarrow u\times w+u\times (w\times u)=u\times v$
$\Rightarrow u\times w+(u\cdot u)w-(u\cdot w)u=u\times v$
$\Rightarrow u\times w+w-(u\cdot w)u=u\times v(2)$
Taking scalar product of (1) with $u$ we get
$u\cdot w+u\cdot (w\times u)=u\cdot v$
$\Rightarrow u\cdot w=u\cdot v$
$[\because u\cdot (w\times u)=0](3)$
Taking scalar product of (1) with $v$ we get
$v\cdot w+v\cdot (w\times u)=v\cdot v$
$\Rightarrow v\cdot w+[vwu]=1$
$\Rightarrow (u\times v)\cdot w=1-v\cdot w(4)$
Taking scalar product of ( 2 ) with $w$ we get
$(u\times w)\cdot w+w\cdot w-(u\cdot w)(u\cdot w)=(u\times v)\cdot w$
$\Rightarrow 0+|w{|}^2-(u\cdot w{)}^2=(u\times v)\cdot w(5)$
$\Rightarrow (u\times v)\cdot w=|w{|}^2-(u\cdot w{)}^2$ .
Taking scalar product of (1) with $w$ we get
$w\cdot w+(w\times u)\cdot w=v\cdot w$
$\Rightarrow |w{|}^2=v\cdot w$
$\Rightarrow |w{|}^2=1-(u\times v)\cdot w[$ using $(4)](6)$
From (5) we get
$(u\times v)\cdot w=|w{|}^2-(u\cdot w{)}^2$
$=1-(u\times v)\cdot w-(u\cdot w{)}^2$
$\Rightarrow 2(u\times v)\cdot w=1-(u\cdot v{)}^2$[Using (3)]
Thus, $|(u\times v)\cdot w|=\frac{1}{2}\left|1-(u\cdot v{)}^2\right|$
$\le \frac{1}{2}\left[\therefore (u\cdot v{)}^2\ge 0\right]$