Spatial distances
Objective. Calculate point-line and point-plane distances.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Spatial distances include the shortest distance from a point to a line and from a point to a plane in three dimensions. The distance from a point P to a plane with equation ax+by+cz=d and normal n=(a,b,c) is |aP₁+bP₂+cP₃−d|/|n|, derived by projecting the vector from any plane point to P onto the unit normal. The distance from a point to a line uses the component of the connecting vector perpendicular to the line's direction, found via the cross product. For example, the distance from the origin to the plane x+2y+2z=9 is |0+0+0−9|/√(1+4+4) = 9/3 = 3.
Worked example
Find the shortest distance from the point (1,2,3) to the plane 2x − y + 2z = 5.
- 1Normal vector is (2,−1,2), with magnitude √(4+1+4)=√9=3
- 2Substitute point into the plane's left-hand side: 2(1)−1(2)+2(3) = 2−2+6 = 6
- 3Distance formula: |6−5|/3
- 4Simplify: 1/3
Answer: 1/3
Concept mastery
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Practice
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Start practisingCommon mistake. Students forget to divide by the magnitude of the normal vector, leaving the distance formula unnormalised and giving an answer with the wrong scale. Always divide the numerator by |n|, the magnitude of the normal vector.
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