Question

A ray of light in incident on a plane mirror along the direction given by the vector a=2i-3j+4k find the unit vector along the reflected ray

Answer 1

Given : The complete question is: ray of light in incident on a plane mirror along the direction given by the vector A=2i-3j+4k and normal to mirror along the direction of  B=3i-6j+3k

To find: The unit vector along the reflected ray.

Solution:

• Now we have given the vectors A=2i-3j+4k and B=3i-6j+3k.

• So, A.B = AB cosФ

            ( 2i-3j+4k ) . ( 3i-6j+3k ) =  √29 x √54 cosФ

            6 + 18 +12 = √1556  ( cos Ф )

            36 / √1556 =  cos Ф

• Now when ray 2i-3j+4k is reflected by the mirror, there is change in direction by the angle will remain same.

• So vector equation of the ray becomes -(2i-3j+4k) and unit vector in the direction of this reflcted ray is:

            n(cap) = -(2i-3j+4k) / √29

Answer:

           So the unit vector along the reflected ray is  -(2i-3j+4k) / √29.

Answer 2

D) None of these

Explanation:

The question is incomplete. Take normal to mirror along the direction of B→ = 3i^ −6j^ ​+2k.

Given: A→ = 2i^ −3j^ ​+4k^.

Solution:

A→.B→ = AB Cosθ

 = (2i^ −3j^ ​+4k^). (3i^ −6j^ ​+2k). Cosθ

We know that Cosθ = 32/ 7√29.

So we  get:

 = √(4 + 9 + 16). √(9 +36 + 4). Cosθ

 The angle between B→ and reflected ray  is same.

So on checking the angles of a unit vector with B in given points, none of the given options matched.

So answer is none of these.

Option D is the answer.