3D Geometry
Reflection in a plane
Grade 12
Question:
<p>The reflection of the point \(\vec{a}\) in the plane \(\vec{r} \cdot \vec{n} = q\) is</p>
<p>\(\vec{a} + \dfrac{(\vec{q} - \vec{a} \cdot \vec{n})}{|\vec{n}|}\)</p>
<p>\(\vec{a} + 2\left(\dfrac{\vec{q} - \vec{a} \cdot \vec{n}}{|\vec{n}|^2}\right)\vec{n}\)</p>
<p>\(\vec{a} + \dfrac{2(\vec{q} + \vec{a} \cdot \vec{n})}{|\vec{n}|}\vec{n}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: The reflection of a point across a plane is found by moving perpendicular to the plane by twice the perpendicular distance from the point to the plane. The perpendicular direction is given by the normal vector to the plane.
Step 1: Let the plane equation be $\vec{r} \cdot \vec{n} = q$, where $\vec{n}$ is the normal vector. Step 2: The perpendicular distance from point $\vec{a}$ to the plane is: $d = \frac{|\vec{a} \cdot \vec{n} - q|}{|\vec{n}|}$ Step 3: The unit normal vector is $\hat{n} = \frac{\vec{n}}{|\vec{n}|}$ Step 4: The reflection of $\vec{a}$ is obtained by moving twice the perpendicular distance along the normal direction away from the plane: $\vec{a}' = \vec{a} - 2\frac{(\vec{a} \cdot \vec{n} - q)}{|\vec{n}|^2}\vec{n}$ Step 5: Simplifying: $\vec{a}' = \vec{a} - 2\frac{(\vec{a} \cdot \vec{n} - q)}{|\vec{n}|^2}\vec{n}$ ∴ Answer: B
Correct Answer: B