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Question
^n1 is the unit vector along incident ray, ^n2 along the normal to plane mirror and ^n3 is the unit vector along reflected ray direction, then which of the following must be true?
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Solution
The correct option is C (^n1×^n2)⋅^n3=0
From the laws of reflection, the incident ray, the reflected ray and the normal to the mirror are always in the same plane.
Here ^n1,^n2 and ^n3 will lie in same plane, for instance xy− plane.
⇒^n1⋅^n2≠0 (∵θ≠90∘) & ^n1⋅^n3≠0 (∵θ≠90∘)
(^n1×^n2) will give a unit vector, which is ⊥r to the plane of (^n1,^n2) i.e ⊥r to the xy− plane.
⇒(^n1×^n2).^n3=0
θ=90∘ between vectors (^n1×^n2) and (^n3). So, all three vectors lie in the same plane.
Hence, option (c) is the correct answer.
Note :- Carefully observe the vectorial representation of ^n1 , ^n2 & ^n3.
From the laws of reflection, the incident ray, the reflected ray and the normal to the mirror are always in the same plane.
Here ^n1,^n2 and ^n3 will lie in same plane, for instance xy− plane.
⇒^n1⋅^n2≠0 (∵θ≠90∘) & ^n1⋅^n3≠0 (∵θ≠90∘)
(^n1×^n2) will give a unit vector, which is ⊥r to the plane of (^n1,^n2) i.e ⊥r to the xy− plane.
⇒(^n1×^n2).^n3=0
θ=90∘ between vectors (^n1×^n2) and (^n3). So, all three vectors lie in the same plane.
Hence, option (c) is the correct answer.
Note :- Carefully observe the vectorial representation of ^n1 , ^n2 & ^n3.
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