Let ABC be a triangle of area 15 √2 and the vectors AB⃗=î+2 ĵ-7 k̂, BC⃗=a î+b ĵ+c k̂ and AC⃗=6 î+d ĵ-2 k̂, d>0. Then the square of the length of the largest side of the triangle ABC is _________.

Solution

Maths sir · 3 steps

Use the area to find d first. Then all three sides are known, so compare their squares.

1. Area to cross product. The area of the triangle is 1/2|AB⃗ × AC⃗|. So |AB⃗ × AC⃗| = 30√2, and its square is 1800.

   AB⃗ × AC⃗ = |î, ĵ, k̂; 1, 2, -7; 6, d, -2| = (7d-4)î - 40ĵ + (d-12)k̂

2. Solve for d. (7d-4)² + 1600 + (d-12)² = 1800 simplifies to 50d² - 80d - 40 = 0, that is 5d² - 8d - 4 = 0. The roots are d = 2 and d = -2/5, and d > 0 gives d = 2.

   Side	Vector	Length squared
   AB	î + 2ĵ - 7k̂	1 + 4 + 49 = 54
   AC	6î + 2ĵ - 2k̂	36 + 4 + 4 = 44
   BC = AC - AB	5î + 0ĵ + 5k̂	25 + 0 + 25 = 50

3. So. The largest side is AB, and the square of its length is 54.

The common slip: 44. Stopping at |AC⃗|² after finding d gives 44. Compare all three sides, because AB is the longest.

The idea to keep. Area of a triangle is 1/2|u⃗ × v⃗| for any two side vectors. Use it to fix the unknown, and remember the third side is the difference AC⃗ - AB⃗.

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