The incident ray,reflected ray,and the outwar | vedclass

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(C) According to the law of reflection,the angle of incidence equals the angle of reflection,and the incident ray,reflected ray,and normal lie in the

Vedclass · Accepted Answer

$\overrightarrow{ b }=\overrightarrow{ a }-2(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$

Vedclass · Answer

(C) According to the law of reflection,the angle of incidence equals the angle of reflection,and the incident ray,reflected ray,and normal lie in the same plane.Let the normal vector be $\overrightarrow{ c }$. The component of the incident ray $\overrightarrow{ a }$ along the normal is $(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$.The component of $\overrightarrow{ a }$ perpendicular to the normal is $\overrightarrow{ a } - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$.Since the reflected ray $\overrightarrow{ b }$ has the same component perpendicular to the normal but the opposite component along the normal,we have:$\overrightarrow{ b } = (\overrightarrow{ a } - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }) - (\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$$\overrightarrow{ b } = \overrightarrow{ a } - 2(\overrightarrow{ a } \cdot \overrightarrow{ c }) \overrightarrow{ c }$

The incident ray,reflected ray,and the outward drawn normal are denoted by the unit vectors $\overrightarrow{ a }, \overrightarrow{ b }$ and $\overrightarrow{ c }$ respectively. Choose the correct relation for these vectors.

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