Vectors 5 Question 9

9. Incident ray is along the unit vector $\hat{\mathbf{v}}$ and the reflected ray is along the unit vector $\hat{\mathbf{w}}$. The normal is along unit vector $\hat{\mathbf{a}}$ outwards. Express $\hat{\mathbf{w}}$ in terms of $\hat{\mathbf{a}}$ and $\hat{\mathbf{v}}$.

$(2005,4$ M)

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Answer:

Correct Answer: 9. $\hat w = \hat v - 2 (\hat a . \hat v ) \hat a $

Solution:

  1. Since, $\hat{\mathbf{v}}$ is unit vector along the incident ray and $\hat{\mathbf{w}}$ is the unit vector along the reflected ray.

Hence, $\hat{\mathbf{a}}$ is a unit vector along the external bisector of $\hat{\mathbf{v}}$ and $\hat{\mathbf{w}}$.

$\therefore \hat{\mathbf{w}}-\hat{\mathbf{v}}=\lambda \hat{\mathbf{a}}$

On squaring both sides, we get

$ \begin{aligned} & \Rightarrow 1+1-\hat{\mathbf{w}} \cdot \hat{\mathbf{v}}=\lambda^{2} \Rightarrow 2-2 \cos 2 \theta=\lambda^{2} \\ & \Rightarrow \quad \lambda=2 \sin \theta \end{aligned} $

where, $2 \theta$ is the angle between $\hat{\mathbf{v}}$ and $\hat{\mathbf{w}}$.

Hence, $\hat{\mathbf{w}}-\hat{\mathbf{v}}=2 \sin \theta \cdot \hat{\mathbf{a}}=2 \cos \left(90^{\circ}-\theta\right) \hat{\mathbf{a}}=-(2 \hat{\mathbf{a}} \cdot \hat{\mathbf{v}}) \hat{\mathbf{a}}$

$\Rightarrow \quad \hat{\mathbf{w}}=\hat{\mathbf{v}}-2(\hat{\mathbf{a}} \cdot \hat{\mathbf{v}}) \hat{\mathbf{a}}$



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