Let A, B, C be three points whose position vectors respectively are
a⃗ = î + 4ĵ + 3k̂
b⃗ = 2î + α ĵ + 4k̂, α ∈ R
c⃗ = 3î - 2ĵ + 5k̂
If α is the smallest positive integer for which a⃗ , b⃗ , c⃗ are noncollinear, then the length of the median, in ΔABC, through A is :

Solution

Maths sir · 4 steps

Find the α that makes the points collinear, step past it, then measure A to the midpoint of BC.

1. When are they collinear. AB⃗ = î + (α - 4)ĵ + k̂ and AC⃗ = 2î - 6ĵ + 2k̂. The points are collinear only if AC⃗ = 2 AB⃗, so 2(α - 4) = -6, giving α = 1.
2. Pick alpha. For every other value of α the points form a triangle. The smallest positive integer other than 1 is α = 2, so B = (2, 2, 4).
3. Midpoint of BC. With C = (3, -2, 5), the midpoint is D = (5/2, 0, 9/2). Then AD⃗ = 3/2î - 4ĵ + 3/2k̂.

   |AD⃗| = √(9/4 + 16 + 9/4) = √(82/4) = √82/2

4. So. The median through A has length √82/2, which is option (a).

The idea to keep. Three points are collinear when AB⃗ and AC⃗ are parallel. The median through a vertex joins it to the midpoint of the opposite side, so find that midpoint first.

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