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NECO 2023 · Paper 1 · Q32

If s=2i−6j\mathbf{s} = 2\mathbf{i} - 6\mathbf{j} and t=−3i+4j\mathbf{t} = -3\mathbf{i} + 4\mathbf{j}, find (t+2s)⋅(t−s)(\mathbf{t} + 2\mathbf{s}) \cdot (\mathbf{t} - \mathbf{s}).

Worked solution (try it first)
  1. t+2s=(−3+4)i+(4−12)j\mathbf t + 2\mathbf s = (-3 + 4)\mathbf i + (4 - 12)\mathbf j
    =i−8j= \mathbf i - 8\mathbf j.
  2. t−s=(−3−2)i+(4+6)j\mathbf t - \mathbf s = (-3 - 2)\mathbf i + (4 + 6)\mathbf j
    =−5i+10j= -5\mathbf i + 10\mathbf j.
  3. The scalar product multiplies the i\mathbf i parts and the j\mathbf j parts, then adds: (1)(−5)+(−8)(10)=−5−80(1)(-5) + (-8)(10) = -5 - 80.
  4. So the value is −85-85, option A.

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