If the components of a⃗=α î+β ĵ+γ k̂ along and perpendicular to b⃗=3 î+ĵ-k̂ respectively, are 16/11(3 î+ĵ-k̂) and 1/11(-4 î-5 ĵ-17 k̂), then α²+β²+γ² is equal to :

Solution

Maths sir · 3 steps

The two components add back up to a⃗ itself, so just add them and read off α, β, γ.

1. Rebuild the vector. Any vector is the sum of its component along b⃗ and its component perpendicular to b⃗. So a⃗ is the sum of the two given vectors.

   a⃗ = 1/11(48î+16ĵ-16k̂) + 1/11(-4î-5ĵ-17k̂) = 1/11(44î+11ĵ-33k̂)

2. Read off components. This simplifies to a⃗ = 4î+ĵ-3k̂, so α = 4, β = 1, γ = -3. As a check, a⃗·b⃗ = 12+1+3 = 16 and |b⃗|² = 11, which matches the parallel part 16/11b⃗.
3. So. α²+β²+γ² = 16+1+9 = 26, which is option (c).

The idea to keep. Parallel part plus perpendicular part always gives back the whole vector. Because the two parts are at right angles, |a⃗|² is also the sum of their squared lengths.

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