Let three vectors ,a⃗=α î+4 ĵ+2 k̂, b⃗=5 î+3 ĵ+4 k̂, c⃗=x î+y ĵ+z k̂ form a triangle such that c⃗=a⃗-b⃗ and the area of the triangle is 5 √6. If α is a positive real number, then |c⃗|² is equal to:

Solution

Maths sir · 4 steps

Half of |a⃗ × b⃗| is the area, so set it equal to 5√6 and find α.

1. Area from cross product. Since c⃗ = a⃗ - b⃗, the triangle has a⃗ and b⃗ as two sides from a common vertex. Its area is 1/2|a⃗ × b⃗| = 5√6, so |a⃗ × b⃗| = 10√6 and |a⃗ × b⃗|² = 600.
2. Cross product. With a⃗ = (α, 4, 2) and b⃗ = (5, 3, 4), a⃗ × b⃗ = 10 î + (10 - 4α) ĵ + (3α - 20) k̂.

   100 + (10 - 4α)² + (3α - 20)² = 600 ⇒ 25α² - 200α = 0

3. Pick alpha. This gives 25α(α - 8) = 0, so α = 0 or α = 8. Since α is positive, α = 8.
4. So. c⃗ = a⃗ - b⃗ = 3î + ĵ - 2k̂, so |c⃗|² = 9 + 1 + 4 = 14. The answer is (a).

The idea to keep. When two vectors from one vertex form a triangle, its area is half the magnitude of their cross product. Square both sides to avoid square roots.

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