Three Dimensional Geometry Question 19
Question: The reflection of the point $ \vec{a} $ in the plane $ \vec{r} $ . $ \vec{n} $ =q is
Options:
A) $ \vec{a}+\frac{(\vec{q}-\vec{a},\cdot \vec{n})}{| {\vec{n}} |} $
B) $ \vec{a}+2( \frac{(\vec{q}-\vec{a},\cdot \vec{n})}{{{| {\vec{n}} |}^{2}}} )\vec{n} $
C) $ \vec{a}+\frac{2(\vec{q}-\vec{a},\cdot \vec{n})}{| {\vec{n}} |}\vec{n} $
D) none of these
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Answer:
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{| {} |}} $
$ \therefore \overrightarrow{OC}=\vec{a}+\frac{(q-\vec{a},\cdot \vec{n})}{\overrightarrow{{{| n |}^{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{{{| n |}^{2}}}}=\frac{\vec{a}+\vec{b}}{2} $ or $ \vec{b}=\vec{a}+2( \frac{q-\vec{a}\cdot \vec{n}}{\overrightarrow{{{| n |}^{2}}}} )\vec{n} $
BETA
