Triangular Plot: Shortest Path, Perpendicular Post and Median
A land surveyor is mapping a triangular plot OAB. From the survey data, OA = 2 units, OB = 3 units, and the angle AOB is 60°. The surveyor writes a and b for the position vectors of A and B relative to O and must answer three practical questions. First, a drainage channel is to run from O to the boundary line AB along the shortest possible path, so the point H where the channel meets AB must be found, together with the length OH, and the surveyor must confirm that H really falls between A and B rather than on an extension of the side. Second, a fence post is to be placed at the point with position vector a + k b, and the post must stand perpendicular to the boundary OB, so the value of k is needed. Third, the surveyor has already marked M, the midpoint of AB, and suspects that OM is longer than OH; this must be checked, together with the area of the plot for the land registry.
1. Find the foot H of the perpendicular from O to the line AB, its position vector, and the length OH.
Read the worked answer
First compute the inner product: a·b = |a||b| cos 60° = 2·3·(1/2) = 3. Also |a|² = 4 and |b|² = 9.
A point of line AB has position vector (1 − t)a + t b. The condition OH ⊥ AB is
((1 − t)a + t b)·(b − a) = 0.
Expand: (1 − t)(a·b − |a|²) + t(|b|² − a·b) = (1 − t)(3 − 4) + t(9 − 3) = −(1 − t) + 6t = −1 + 7t.
Setting this to zero gives t = 1/7, so
oh = (6/7)a + (1/7)b.
Since 0 ≤ 1/7 ≤ 1, H lies on the segment AB, not on an extension - the surveyor's check passes.
Length: |oh|² = (36|a|² + 12(a·b) + |b|²)/49 = (144 + 36 + 9)/49 = 189/49 = 27/7, so OH = √(27/7) = 3√21/7 ≈ 1.96 units.
2. Find k such that the point with position vector a + k b gives a post perpendicular to OB.
Read the worked answer
The direction of OB is b, so perpendicularity to OB means the vector from O to the post is orthogonal to b:
(a + k b)·b = a·b + k|b|² = 3 + 9k = 0,
hence k = −1/3.
The negative sign is not an error: it says the post lies on the side of O opposite to B along the direction of b, while still satisfying the perpendicularity requirement relative to the line OB. A frequent mistake is to solve (a + k b)·a = 0, which imposes perpendicularity to OA instead; reading which line the post must be perpendicular to is the whole judgement in this item.
3. Compare OM with OH and compute the area of the plot.
Read the worked answer
M is the midpoint of AB, so om = (a + b)/2 and
|om|² = (|a|² + 2a·b + |b|²)/4 = (4 + 6 + 9)/4 = 19/4,
so OM = √19/2 ≈ 2.18, versus OH ≈ 1.96. The suspicion is correct: OM > OH.
This is the expected outcome, not a coincidence. OH is the distance from O to the line AB, the minimum over all points of that line, and M corresponds to t = 1/2 ≠ 1/7 = t_H, so M is a different point of the line and must be farther from O. Had the two lengths come out equal, it would have signalled an arithmetic slip.
Area of triangle OAB:
Area = (1/2)|a||b| sin 60° = (1/2)(2)(3)(√3/2) = 3√3/2 ≈ 2.60 square units.
A vector-only check: |a|²|b|² − (a·b)² = 36 − 9 = 27, and half the square root gives 3√3/2, confirming the result without using the angle explicitly.