Let a⃗=9 î-13 ĵ+25 k̂, b⃗=3 î+7 ĵ-13 k̂ and c⃗=17 î-2 ĵ+k̂ be three given vectors. If r⃗ is a vector such that r⃗ × a⃗=(b⃗+c⃗) × a⃗ and r⃗ ·(b⃗-c⃗)=0, then (|593 r⃗+67 a⃗|²)/(593)² is equal to __________.

Solution

Maths sir · 4 steps

Equal cross products with a⃗ mean r⃗=b⃗+c⃗+ta⃗; then find t.

1. Read the cross condition. Move everything to one side: (r⃗-b⃗-c⃗) × a⃗ = 0⃗. So r⃗-b⃗-c⃗ is parallel to a⃗, which gives r⃗ = b⃗+c⃗+ta⃗ for some scalar t.
2. Use the dot condition. Put this r⃗ into r⃗·(b⃗-c⃗)=0. You get (b⃗+c⃗)·(b⃗-c⃗) + t a⃗·(b⃗-c⃗) = 0, and the first term is |b⃗|²-|c⃗|² = 227-294 = -67.
3. Find t. With b⃗-c⃗ = -14î+9ĵ-14k̂, the product is a⃗·(b⃗-c⃗) = -126-117-350 = -593. So -67-593t=0, which gives t=-67/593.

   593r⃗+67a⃗ = 593(b⃗+c⃗)

4. So. The expression equals |b⃗+c⃗|². With b⃗+c⃗ = 20î+5ĵ-12k̂, this is 400+25+144 = 569.

The idea to keep. Whenever you see r⃗×a⃗ = p⃗×a⃗, write r⃗ = p⃗+ta⃗ at once. Then use the remaining dot condition to find t.

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