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MathematicsMadrid 2026Question 3Free lesson

Reflection, distance and orthogonal plane

3D geometry: why the plane's normal vector decides the perpendicular, the distance and the orthogonal plane

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Problem statement

Given the plane $\pi: 2x+2y-z=13$ and the line $$r\equiv\dfrac{x-2}{1}=\dfrac{y}{1}=\dfrac{z}{4}.$$

a) (1 mark) Find the reflection (mirror image) of the point $P(2,0,0)$ in the plane $\pi$.

b) (1 mark) Answer only one of the following two parts:

b1) Find the distance between the plane $\pi$ and the line $r$.

b2) Find an equation of the plane that contains the line $r$ and is orthogonal to the plane $\pi$.

Source: Madrid university entrance exam (PAU, Spain) · 2025-2026 · Ordinary session · Mathematics · Question 3. Official exam set by the public universities of the Community of Madrid, to whom it belongs; translated from Spanish and reproduced with its source in order to explain it for teaching purposes.

What this lesson answers

  1. What do the equations tell us before any calculation?
  2. How do we construct the reflection of a point in a plane?
  3. Does it make sense to talk about the distance between this line and this plane?
  4. Why is one point of the line enough to measure the distance?
  5. What does a plane need in order to be determined?
  6. Another route for each part: does it give the same?
  7. The answers and how to check them
  8. Common mistakes to avoid
  9. Exam tips
  10. For top marks

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