HCJC 1999 P1 Q19
Points A, B, C D have position vectors -3i + 4j, 3i + pj, 2i + 3j and i - 2j respectively.
(i) Given that AB·AD = 0, find the value of p.
(ii) Show that ∡BAC = ∡DAC.
(iii) Show that the points B, C, D are collinear and write down the vector equation of the line l through these three points.
(iv) Find the position vector of N, the foot of the perpendicular from the point E (5, 18, -4) to the line l. Find also the shortest distance from E to the line l.
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Answer:
(i) AB = OB - OA = 6i + (p-4)j
AD = OD - OA = 4i - 6j
AB·AD =0
24 - (p-4)*6 = 0
24 - 6p + 24 = 0
p = 8
(ii) AC = OC - OA = 5i - j
AB = 6i + 4j
AD = 4i - 6j
Since cos-1 ∡BAC = cos-1 ∡DAC
=> ∡BAC = ∡DAC (shown)
(iii) BC = OC - OB = -i - 5j
BD = OD - OB = -2i - 10j = 2 BC
Thus, B, C and D are collinear (shown)
Line l
(iv) Since N is on l, let N =
Hence,
Since
2 - λ + 50 - 25λ = 0
26λ = 52
λ = 2
Hence,
