Let a⃗ = î + 2ĵ + λ k̂,b⃗ = 3î - 5ĵ - λ k̂,a⃗ . c⃗ = 7,2b⃗ . c⃗ + 43 = 0,a⃗ × c⃗ = b⃗ × c⃗ . Then | a⃗ . b⃗ | is equal to :

Solution

Maths sir · 4 steps

Bring both cross products to one side and c⃗ becomes a multiple of a⃗ - b⃗.

1. Parallel vectors. From a⃗ × c⃗ = b⃗ × c⃗ we get (a⃗ - b⃗) × c⃗ = 0⃗. So c⃗ = t(a⃗ - b⃗) for some scalar t, where a⃗ - b⃗ = -2î + 7ĵ + 2λ k̂.
2. Two dot conditions. a⃗ · c⃗ = t(-2 + 14 + 2λ²) = t(12 + 2λ²) = 7. Also b⃗ · c⃗ = -43/2, so (a⃗ - b⃗) · c⃗ = 7 + 43/2 = 57/2. This gives t(4 + 49 + 4λ²) = t(53 + 4λ²) = 57/2.

   (53 + 4λ²)/(12 + 2λ²) = 57/14 ⇒ 742 + 56λ² = 684 + 114λ² ⇒ λ² = 1

3. Find the dot product. a⃗ · b⃗ = (1)(3) + (2)(-5) + (λ)(-λ) = 3 - 10 - λ² = -8. As a check, t = 7/14 = 1/2, and 1/2 × 57 = 57/2 fits the second condition.
4. So. |a⃗ · b⃗| = |-8| = 8.

The common slip: 6. Taking the k̂ product as +λ² instead of λ × (-λ) = -λ² gives 3 - 10 + 1 = -6, so a modulus of 6 instead of 8.

The idea to keep. If a⃗ × c⃗ = b⃗ × c⃗, then c⃗ is a multiple of a⃗ - b⃗. Divide the two dot conditions and that multiple cancels out.

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