Let a⃗=î+2 ĵ+3 k̂, b⃗=2 î+3 ĵ-5 k̂ and c⃗=3 î-ĵ+λ k̂ be three vectors. Let r⃗ be a unit vector along b⃗+c⃗. If r⃗ · a⃗=3, then 3 λ is equal to:

Solution

Maths sir · 5 steps

A unit vector hides a modulus in the denominator, so write r⃗ out in full and then square.

1. Build the vector. b⃗+c⃗ = 5î+2ĵ+(λ-5)k̂. Its modulus is √(25+4+(λ-5)²) = √(29+(λ-5)²), and r⃗ is this vector divided by that modulus.
2. Dot with a. (b⃗+c⃗)·a⃗ = 5(1)+2(2)+3(λ-5) = 3λ-6. So the condition r⃗·a⃗=3 becomes the equation below.

   (3λ-6)/√(29+(λ-5)²) = 3

3. Square and solve. (3λ-6)² = 9(29+(λ-5)²) gives 9λ²-36λ+36 = 9λ²-90λ+486. The λ² terms cancel, leaving 54λ = 450, so λ = 25/3.
4. Check the sign. Squaring can bring in a false root, so check it. With λ = 25/3, the top is 3λ-6 = 19 > 0 and the modulus is √(29+100/9) = 19/3. That gives 19 ÷ 19/3 = 3, as required.
5. So. 3λ = 25, which is option (b).

The idea to keep. A unit vector along v⃗ is v⃗/|v⃗|. Write the modulus in, square to clear the root, and then check the sign of the dot product you squared.

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