Let the position vectors of the vertices A, B and C of a triangle be 2 î+2 ĵ+k̂, î+2 ĵ+2 k̂ and 2 î+ĵ+2 k̂ respectively. Let l₁, l₂ and l₃ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides AB, BC and CA respectively, then l₁²+l₂²+l₃² equals:

Solution

Maths sir · 4 steps

Check the side lengths first: all three are √2, so the orthocentre is just the centroid.

1. Find the sides. AB⃗ = -î+k̂, BC⃗ = î-ĵ and CA⃗ = ĵ-k̂. Each has length √(1+1) = √2, so the triangle is equilateral.
2. Locate the orthocentre. In an equilateral triangle the orthocentre, centroid and incentre are the same point. So H = (A+B+C)/3 = 5/3(î+ĵ+k̂). Its perpendicular distance to each side equals the inradius.

   l₁ = l₂ = l₃ = r = a/2√3 = √2/2√3 = 1/√6

3. Quick check. The midpoint of AB is (3/2, 2, 3/2). The vector from it to H is (1/6, -1/3, 1/6), which is perpendicular to AB. Its squared length is 1/36+4/36+1/36 = 1/6, the same as l₁².
4. So. l₁²+l₂²+l₃² = 3 × 1/6 = 1/2, which is option (d).

The idea to keep. Before hunting for an orthocentre with dot products, compute the three side lengths. If the triangle is equilateral, every centre is the centroid and every perpendicular to a side is the inradius a/2√3.

Answer from the NTA final answer key. Solution by PrepWiser.