Given: A(1, 8, 4)
Line segment joining B(0, -1, 3) and C(2, -3, -1) is
BC = 2i – 2j – 4k
Let the foot of the perpendicular be R then,
As R lies on the line having point B and parallel to BC,
So, R = (0, -1, 3) + a(2, -2, -4)
R(2a, -1-2a, 3-4a)
The line segment AR is
AR = (2a-1)i + (-1-2a-8)j + (3-4x-4)k
As the lines AR and BC are perpendicular thus, (as R is the foot of the perpendicular on BC)
AR.BC = 0
2(2a-1) + (-2)(-9-2a) + (-4)(-1-4a) = 0
24a + 20 = 0
a = \(-\frac{5}{6}\)
Substituting a in R we get,
\(R(-\frac{5}{3},\frac{2}{3},\frac{19}{3})\)
