Let a⃗=6 î+9 ĵ+12 k̂, b⃗=α î+11 ĵ-2 k̂ and c⃗ be vectors such that a⃗ × c⃗=a⃗ × b⃗. If
a⃗ · c⃗=-12, c⃗ ·(î-2 ĵ+k̂)=5, then c⃗ ·(î+ĵ+k̂) is equal to _______________.

Solution

Maths sir · 5 steps

Equal cross products with a⃗ mean c⃗=b⃗+λa⃗, so start there.

1. Remove the cross product. From a⃗×c⃗=a⃗×b⃗ we get a⃗×(c⃗-b⃗)=0⃗. So c⃗-b⃗ is parallel to a⃗, and c⃗=b⃗+λa⃗ for some scalar λ.
2. Find alpha first. Dot with î-2ĵ+k̂. The a⃗ part gives 6-18+12=0, so λ drops out. That leaves α-22-2=5, so α=29.
3. Find lambda. Now a⃗·b⃗=6(29)+99-24=249 and |a⃗|²=36+81+144=261. Then a⃗·c⃗=-12 gives the equation below.

   249+261λ=-12 ⇒ λ=-1

4. Build c. c⃗=b⃗-a⃗=23î+2ĵ-14k̂. Check: a⃗·c⃗=138+18-168=-12 and 23-4-14=5, both correct.
5. So. c⃗·(î+ĵ+k̂)=23+2-14= 11.

The common slip: 65. Getting the sign of λ wrong and using c⃗=b⃗+a⃗ gives 65, not 11.

The idea to keep. Whenever a⃗×c⃗=a⃗×b⃗, write c⃗=b⃗+λa⃗. Then look for a condition perpendicular to a⃗: it kills λ and gives the other unknown.

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