Vectors 3 Question 27
27. If $\overrightarrow{\mathbf{u}}, \overrightarrow{\mathbf{v}}, \overrightarrow{\mathbf{w}}$ are three non-coplanar unit vectors and $\alpha, \beta, \gamma$ are the angles between $\overrightarrow{\mathbf{u}}$ and $\overrightarrow{\mathbf{v}}, \overrightarrow{\mathbf{v}}$ and $\overrightarrow{\mathbf{w}}, \overrightarrow{\mathbf{w}}$ and $\overrightarrow{\mathbf{u}}$ respectively and $\overrightarrow{\mathbf{x}}, \overrightarrow{\mathbf{y}}, \overrightarrow{\mathbf{z}}$ are unit vectors along the bisectors of the angles $\alpha, \beta, \gamma$ respectively. Prove that
$[\overrightarrow{\mathbf{x}} \times \overrightarrow{\mathbf{y}} \overrightarrow{\mathbf{y}} \times \overrightarrow{\mathbf{z}} \overrightarrow{\mathbf{z}} \times \overrightarrow{\mathbf{x}}]=\frac{1}{16}[\overrightarrow{\mathbf{u}} \overrightarrow{\mathbf{v}} \overrightarrow{\mathbf{w}}]^{2} \sec ^{2} \frac{\alpha}{2} \sec ^{2} \frac{\beta}{2} \sec ^{2} \frac{\gamma}{2}$
(2003, 4M)
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Answer:
Correct Answer: 27. $\frac{146}{17}$
Solution:
- $\quad \overrightarrow{\mathbf{x}}=\frac{\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}}|}=\frac{1}{2} \sec \frac{\alpha}{2}(\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}})$
$$ \begin{array}{r} \overrightarrow{\mathbf{y}}=\frac{\overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}}|}=\frac{1}{2} \sec \frac{\beta}{2}(\overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}}) \\ \overrightarrow{\mathbf{z}}=\frac{\overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{u}}}{|\overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{u}}|}=\frac{1}{2} \sec \frac{\gamma}{2}(\overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{u}}) \end{array} $$
Since, $\quad[\overrightarrow{\mathbf{x}} \times \overrightarrow{\mathbf{y}} \overrightarrow{\mathbf{y}} \times \overrightarrow{\mathbf{z}} \overrightarrow{\mathbf{z}} \times \overrightarrow{\mathbf{x}}]=[\overrightarrow{\mathbf{x}} \overrightarrow{\mathbf{y}} \overrightarrow{\mathbf{z}}]^{2}$
[from Eq. (i)]
$$ =\frac{1}{64} \sec ^{2} \frac{\alpha}{2} \cdot \sec ^{2} \frac{\beta}{2} \cdot \sec ^{2} \frac{\gamma}{2}[\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}} \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}} \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{u}}]^{2} $$
and $[\overrightarrow{\mathbf{u}}+\overrightarrow{\mathbf{v}} \overrightarrow{\mathbf{v}}+\overrightarrow{\mathbf{w}} \overrightarrow{\mathbf{w}}+\overrightarrow{\mathbf{u}}]=2[\overrightarrow{\mathbf{u}} \overrightarrow{\mathbf{v}} \overrightarrow{\mathbf{w}}]$
$\therefore[\overrightarrow{\mathbf{x}} \times \overrightarrow{\mathbf{y}} \overrightarrow{\mathbf{y}} \times \overrightarrow{\mathbf{z}} \overrightarrow{\mathbf{z}} \times \overrightarrow{\mathbf{x}}]$
$$ \begin{aligned} & =\frac{1}{64} \sec ^{2} \frac{\alpha}{2} \cdot \sec ^{2} \frac{\beta}{2} \cdot \sec ^{2} \frac{\gamma}{2} \cdot 4[\overrightarrow{\mathbf{u}} \overrightarrow{\mathbf{v}} \overrightarrow{\mathbf{w}}]^{2} \\ & =\frac{1}{16} \cdot[\overrightarrow{\mathbf{u}} \overrightarrow{\mathbf{v}} \overrightarrow{\mathbf{w}}]^{2} \sec ^{2} \frac{\alpha}{2} \cdot \sec ^{2} \frac{\beta}{2} \cdot \sec ^{2} \frac{\gamma}{2} \end{aligned} $$
