Lesson 7 of 9 · 6 min
Foot of a perpendicular JEE adds
NCERT §11.3–11.5, JEE extension
A light sensor is fixed to the ceiling at S(6, 3, 6). Which point of the first beam is nearest to it, and how far is it?
The lesson in notes
In short
To drop a perpendicular from a point S to the line r = a + λb, write the general point N = a + λb and demand SN · b = 0. This one equation in λ fixes the foot N.
Equivalently λ = (s − a)·b / |b|², where s is the position vector of S.
The distance from S to the line is |SN|, which also equals |b × (s − a)|/|b|, the same expression as for parallel lines.
The image (mirror point) of S in the line is S′ = 2N − S, since the foot N is the mid-point of SS′.
Worked: for S(6, 3, 6) and the line through (1, 2, 0) along 2î + 3ĵ + 6k̂, (s − a)·b = 10 + 3 + 36 = 49 and |b|² = 49, so λ = 1 and N = (3, 5, 6). Then |SN| = √(9 + 4) = √13 and the image is (0, 7, 6).
The ends of the shortest segment between skew lines come the same way: take a general point on each line and make the joining vector perpendicular to both directions (two equations, two parameters).