$\vec{u}, \vec{v}, \vec{w}$ are three unit vectors. Let $\vec{p}=\vec{u}+\vec{v}+\vec{w}, \vec{q}$ $=\vec{u}…
$\vec{u}, \vec{v}, \vec{w}$ are three unit vectors. Let $\vec{p}=\vec{u}+\vec{v}+\vec{w}, \vec{q}$ $=\vec{u} \times(\vec{v} \times \vec{w})$.If $\vec{p} \cdot \vec{u}=\frac{3}{2}, \vec{p} \cdot \vec{v}=\frac{7}{4},|\vec{p}|=2$ and $\vec{v}=\mathrm{K} \vec{q}$, then $K=$
$-1$
$2$
$3$
$-2$
Solution
Given that $|\vec{u}|=|\vec{v}|=|\vec{w}|=1$
$\therefore \vec{p} \cdot \vec{u}=(\vec{u}+\vec{v}+\vec{w}) \cdot \vec{u}=\frac{3}{2} \Rightarrow|\vec{u}|^2+\vec{u} \cdot \vec{v}+\vec{w} \cdot \vec{u}=\frac{3}{2}$
$\Rightarrow \vec{u} \cdot \vec{v}+\vec{w} \cdot \vec{u}=\frac{3}{2}-1=\frac{1}{2}$ ....(i)
$\vec{p} \cdot \vec{v}=(\vec{u}+\vec{v}+\vec{w}) \cdot \vec{v}=\frac{7}{4} \Rightarrow \vec{u} \cdot \vec{v}+|\vec{v}|^2+\vec{v} \cdot \vec{w}=\frac{7}{4}$
$\Rightarrow \vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}=\frac{7}{4}-1=\frac{3}{4}$ ....(ii)
$|\vec{p}|^2=|\vec{u}+\vec{v}+\vec{w}|^2$
$\Rightarrow 4=|\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})$
$\Rightarrow \frac{1}{2}=\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}$ ....(iii)
On solving (i), (ii) and (iii), we get
$\vec{u} \cdot \vec{v}=\frac{3}{4}, \vec{v} \cdot \vec{w}=0$ and $\vec{w} \cdot \vec{u}=-\frac{1}{4}$
$\therefore \vec{v}=k \vec{q}=k[\vec{u} \times(\vec{v} \times \vec{w})]=k[(\vec{u} \cdot \vec{w}) \vec{v}-(\vec{u} \cdot \vec{v}) \vec{w}]$
$\vec{v}=k\left(-\frac{1}{4} \vec{v}-\frac{3}{4} \vec{w}\right) \Rightarrow 4 \vec{v}=-k \vec{v}-3 k \vec{w}$
$(4+k) \vec{v}=-3 k \vec{w} \Rightarrow|4+k||\vec{v}|=|-3 k||\vec{w}|$
$\therefore|4+k|=|3 k| \Rightarrow 4+k=3 k \Rightarrow k=2$