question_answer

The reflection of the point \[\vec{a}\] in the plane \[\vec{r}\].\[\vec{n}\] =q  is

A) \[\vec{a}+\frac{(\vec{q}-\vec{a}\,\cdot \vec{n})}{\left| {\vec{n}} \right|}\]          

B) \[\vec{a}+2\left( \frac{(\vec{q}-\vec{a}\,\cdot \vec{n})}{{{\left| {\vec{n}} \right|}^{2}}} \right)\vec{n}\]

C) \[\vec{a}+\frac{2(\vec{q}-\vec{a}\,\cdot \vec{n})}{\left| {\vec{n}} \right|}\vec{n}\]

D) none of these

Correct Answer: B

Solution :

[b] Given plane is \[\vec{r}\cdot \vec{n}=q\]                    ...(i) Let the image of A \[(\vec{a})\] in the plane be B \[(\vec{b})\]. Equation of AC is \[\vec{r}=\vec{a}+\lambda \vec{n}\] (\[\therefore \]AC is normal to the plane)                ...(ii) Solving (i) and (ii). We get \[(\vec{a}+\lambda \vec{n})\cdot \vec{n}=q\] Or \[\lambda =\frac{q-\vec{a}\cdot \vec{n}}{\overrightarrow{\left| {} \right|}}\] \[\therefore \overrightarrow{OC}=\vec{a}+\frac{(q-\vec{a}\,\cdot \vec{n})}{\overrightarrow{{{\left| n \right|}^{2}}}}\cdot \vec{n}\] But \[\overrightarrow{OC}=\frac{\vec{a}+\vec{b}}{2}\] \[\therefore \,\,\vec{a}+\frac{(q-\vec{a}\,\cdot \vec{n})\vec{n}}{\overrightarrow{{{\left| n \right|}^{2}}}}=\frac{\vec{a}+\vec{b}}{2}\] or \[\vec{b}=\vec{a}+2\left( \frac{q-\vec{a}\cdot \vec{n}}{\overrightarrow{{{\left| n \right|}^{2}}}} \right)\vec{n}\]