A lvl H2 Maths: Vectors

Question from http://www.sgforums.com/forums/2297/topics/407847

Relative to an origin O, the position vectors of points A and B are a and b respectively. Givent that angle AOB is 90deg, show that the position vector of the foot of the perpendicular from O to AB is

a + (|a|² / (|a|² + |b|²)) (b - a)

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Answer:

Using the concept of vector projection, (i.e. AF = |AF| * unit vector AB),

OF = OA + AF = a + |a| cos θ * (b - a) / |b - a|, where θ = angle between OA and BA
= a + |a| [ {a.(a - b)} / {|a| |a - b| }] * (b - a) / |b - a|
= a + {a.a - a.b} / (|a|² + |b|²) * (b - a)
= a + |a|² / (|a|² + |b|²) * (b - a) since a.a = |a|² and a.b = 0

(shown)

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