Question

A plane is moving with velocity $4\mathop i + 5\mathop j + 8\mathop k$. A point object in front of the mirror moves with a velocity $3\mathop i + 4\mathop j + 5\mathop k$. Here $\mathop k$ is along the normal to the plane mirror and facing towards the object. The velocity of the image is?
A.$- 3\mathop i - 4\mathop j + 5\mathop k$
B.$3\mathop i + 4\mathop j + 11\mathop k$
C.$- 4\mathop i + 5\mathop j + 11\mathop k$
D.$7\mathop i + 9\mathop j + 11\mathop k$

Answer 1

Here, we have to use the concept of the perpendicular component and parallel component of the velocity of the image on a plane mirror to find out the vector of the velocity of the image. So by simply applying the formula of the perpendicular component and parallel component of the velocity of a vector, we will get the velocity of the image.

Complete step-by-step answer:
Let $\mathop {{{\rm{V}}_{\rm{p}}}}\limits^ \to$be the velocity of the plane. Therefore,$\mathop {{{\rm{V}}_{\rm{p}}}}\limits^ \to {\rm{ = }}4\mathop i + 5\mathop j + 8\mathop k$
Let $\mathop {{{\rm{V}}_{\rm{o}}}}\limits^ \to$ be the velocity of the object. Therefore, $\mathop {{{\rm{V}}_{\rm{o}}}}\limits^ \to {\rm{ = }}3\mathop i + 4\mathop j + 5\mathop k$
Let $\mathop {{{\rm{V}}_{\rm{i}}}}\limits^ \to$be the velocity of the image which we have to find.
So by using the concept of the perpendicular component of the velocity of an image on a plane mirror we will get the perpendicular component of the velocity of the image. Therefore, we know that the perpendicular component of the velocity of a plane is equal to the average of the perpendicular component of the velocity of the image and the object. So, we get
${\left( {\mathop {{{\rm{V}}_{\rm{p}}}}\limits^ \to } \right)_ \bot } = \dfrac{{{{\left( {\mathop {{{\rm{V}}_{\rm{i}}}}\limits^ \to } \right)}_ \bot } + {{\left( {\mathop {{{\rm{V}}_{\rm{o}}}}\limits^ \to } \right)}_ \bot }}}{2}$
So by solving this, we will get the perpendicular i.e. z component of the velocity of the image.
$\Rightarrow 8 = \dfrac{{{{\left( {\mathop {{{\rm{V}}_{\rm{i}}}}\limits^ \to } \right)}_ \bot } + 5}}{2}$
$\Rightarrow {\left( {\mathop {{{\rm{V}}_{\rm{i}}}}\limits^ \to } \right)_ \bot } = 11$
So, $11\mathop k$ is the z component of the velocity of the image.
Now we have to use the concept of the parallel component of the velocity of the image on a plane mirror then we will get the parallel components of the velocity of the image. Therefore, we know that the parallel component of the velocity of the image always remains equal to the parallel component of the velocity of the object itself. So, we get
${\left( {\mathop {{{\rm{V}}_{\rm{o}}}}\limits^ \to } \right)_\parallel } = {\left( {\mathop {{{\rm{V}}_{\rm{i}}}}\limits^ \to } \right)_\parallel }$
Therefore, the x and y component of the image will remain the same as that of the x and y component of the velocity of the object i.e. $3\mathop i$ and $4\mathop j$ respectively.
Therefore the velocity of the image is $\mathop {{{\rm{V}}_{\rm{i}}}}\limits^ \to {\rm{ = }}3\mathop i + 4\mathop j + 11\mathop k$
Hence, option B is the correct option.

Note: We should know that Vectors have three components i.e. x component, y component and z component and all the three components of the vectors are perpendicular to each other. A mirror is a plane that has only x and y components. That is why the x and y component of the image remained the same as that of the x and y component of the object but the perpendicular component of the image varied. The concept of the velocity of the image on a plane mirror states that the parallel components of the image remain the same as that of the object velocity components and the perpendicular component of the velocity of the image is different from the perpendicular component of the velocity of the object.