JEE Main 2021 · 25 February · morning · Single correct · +4 / −1
In an octagon ABCDEFGH of equal side, what is the sum of
AB⃗+AC⃗+AD⃗+AE⃗+AF⃗+AG⃗+AH⃗,
if, AO⃗=2î+3ĵ-4k̂
- (a) -16î-24ĵ+32k̂
- (b) 16î+24ĵ-32k̂
- (c) 16î+24ĵ+32k̂
- (d) 16î-24ĵ+32k̂
Answer and solution
Answer (b) 16î+24ĵ-32k̂
Solution
Physics sir · 4 steps
Break every vector at the centre O: each AX becomes AO plus OX, and the OX vectors almost cancel.
- Route through O. Write each vector as a trip through the centre: AX⃗ = AO⃗ + OX⃗ for X = B, C, D, E, F, G, H. There are seven such terms, so AO⃗ appears 7 times.
Sum = 7 AO⃗ + (OB⃗+OC⃗+OD⃗+OE⃗+OF⃗+OG⃗+OH⃗)
- Use the symmetry. In a regular octagon the eight vectors from the centre to the vertices add to zero, because each one is cancelled by its opposite. So the seven in the bracket add to -OA⃗ = AO⃗.
- Combine. The sum is 7 AO⃗ + AO⃗ = 8 AO⃗ = 8(2î+3ĵ-4k̂).
- So. The sum is 16î+24ĵ-32k̂, which is option (b).
The idea to keep. For sums of vectors from one vertex of a regular polygon, go through the centre. The centre-to-vertex vectors add to zero, so an n-sided polygon gives n AO⃗.
Answer from the NTA final answer key. Solution by PrepWiser.