Vectors 3 Question 29
29. Let $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}$ be unit vectors. If $\overrightarrow{\mathbf{w}}$ is a vector such that $\quad \overrightarrow{\mathbf{w}}+(\overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}})=\overrightarrow{\mathbf{v}}$, then prove that $|(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}| \leq \frac{1}{2}$ and that the equality holds if and only if $\overrightarrow{\mathbf{u}}$ is perpendicular to $\overrightarrow{\mathbf{v}}$.
$(1999,10 M)$
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Solution:
- Given equation is $\overrightarrow{\mathbf{w}}+(\overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}})=\overrightarrow{\mathbf{v}}$
Taking cross product with $\overrightarrow{\mathbf{u}}$, we get
$$ \begin{array}{rr} & \overrightarrow{\mathbf{u}} \times[\overrightarrow{\mathbf{w}}+(\overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}})]=\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}} \\ \Rightarrow & \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{u}} \times(\overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}})=\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}} \\ \Rightarrow & \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{w}}+(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{u}}) \overrightarrow{\mathbf{w}}-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}}) \overrightarrow{\mathbf{u}}=\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}} \\ \Rightarrow & \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}}-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}}) \overrightarrow{\mathbf{u}}=\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}} \end{array} $$
Now, taking dot product of Eq. (i) with $\overrightarrow{\mathbf{u}}$, we get
$$ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{u}} \cdot(\overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}})=\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} $$
$\Rightarrow \quad \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}}=\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} \quad[\because \overrightarrow{\mathbf{u}} \cdot(\overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}})=0]$
Now, taking dot product of Eq. (i) with $\overrightarrow{\mathbf{u}}$, we get $\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{v}} \cdot(\overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}})=\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{v}}$
$\Rightarrow \overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+[\overrightarrow{\mathbf{v}} \overrightarrow{\mathbf{w}} \overrightarrow{\mathbf{u}}]=1 \Rightarrow \overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+[\overrightarrow{\mathbf{v}} \overrightarrow{\mathbf{w}} \overrightarrow{\mathbf{u}}]-1=0$
$\Rightarrow-(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}-\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}+1=0$
$\Rightarrow \quad 1-\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}=(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}$
Taking dot product of Eq. (ii) with $\overrightarrow{\mathbf{w}}$, we get $(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{w}}) \cdot \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{w}}-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}})(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}})=(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}$
$$ \begin{array}{ll} \Rightarrow & 0+|\overrightarrow{\mathbf{w}}|^{2}-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}})^{2}=(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}} \\ \Rightarrow & (\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}=|\overrightarrow{\mathbf{w}}|^{2}-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}})^{2} \end{array} $$
Taking dot product of Eq. (i) with $\overrightarrow{\mathbf{w}}$, we get
$$ \begin{aligned} & \overrightarrow{\mathbf{w}} \cdot \overrightarrow{\mathbf{w}}+(\overrightarrow{\mathbf{w}} \times \overrightarrow{\mathbf{u}}) \cdot \overrightarrow{\mathbf{w}}=\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}} \\ \Rightarrow & |\overrightarrow{\mathbf{w}}|^{2}+0=\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}} \\ \Rightarrow & |\overrightarrow{\mathbf{w}}|^{2}=1-(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}} \quad \text { [from Eq. (iv)] } \end{aligned} $$
Again, from Eq. (v), we get
$(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}+|\overrightarrow{\mathbf{w}}|^{2}-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}})^{2}=1-(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}})^{2}$
$\Rightarrow \quad 2(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}=1-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}})^{2}$
[from Eq. (iii)]
$$ \Rightarrow \quad|(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}}|=\frac{1}{2}\left|1-(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}})^{2}\right| \leq \frac{1}{2} \quad\left[\because(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}})^{2} \geq 0\right] $$
The equality holds if and only if $\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}=0$ or iff $\overrightarrow{\mathbf{u}}$ is perpendicular to $\overrightarrow{\mathbf{v}}$.
