Let v⃗=α î+2 ĵ-3 k̂, w⃗=2 α î+ĵ-k̂ and u⃗ be a vector such that |u⃗|=α>0. If the minimum value of the scalar triple product [ u⃗, v⃗, w⃗ ] is -α √3401, and |u⃗ · î|²=m/n where m and n are coprime natural numbers, then m+n is equal to ____________.

Solution

Maths sir · 4 steps

Write the box product as a dot product, then point u⃗ opposite to v⃗×w⃗.

1. Cross product first. With v⃗=(α, 2, -3) and w⃗=(2α, 1, -1), we get v⃗×w⃗=(1, -5α, -3α). Its length is √(1+25α²+9α²)=√(1+34α²).
2. Least value. [u⃗ v⃗ w⃗]=u⃗·(v⃗×w⃗)=|u⃗| |v⃗×w⃗|cos θ. It is least when cos θ=-1, that is, when u⃗ is exactly opposite to v⃗×w⃗.

   -α√(1+34α²)=-α√3401 ⇒ 34α²=3400 ⇒ α=10

3. Find the vector u. Now v⃗×w⃗=(1,-50,-30) with length √3401. So u⃗=-10 (1,-50,-30)/√3401, which gives u⃗·î=-10/√3401.
4. So. |u⃗·î|²=100/3401. Since 3401=19× 179 shares no factor with 100, m=100, n=3401 and m+n= 3501.

The common slip: 3402. Treating u⃗ as a unit vector and forgetting |u⃗|=10 gives 1/3401 and so 3402, not 3501.

The idea to keep. A box product [u⃗ v⃗ w⃗] with |u⃗| fixed ranges between ±|u⃗| |v⃗×w⃗|. The extremes come when u⃗ is parallel or antiparallel to v⃗×w⃗.

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