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. $\leq \frac{1}{3}$
• B. $\leq \frac{1}{2}$
• C. $ > \frac{1}{3}$
• D. $\geq \frac{1}{2}$

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 $
$[\because \quad u \cdot(w \times u)=0]\,\,\,(3)$
Taking scalar product of (1) with $v$ we get
$v \cdot w+v \cdot(w \times u)=v \cdot v $
$\Rightarrow v \cdot w+[v w u]=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, $|(u \times v) \cdot w|=\frac{1}{2}\left|1-(u \cdot v)^{2}\right|$
$\leq \frac{1}{2}\left[\therefore (u \cdot v)^{2} \geq 0\right]$