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NDA 2019 Maths · Mathematics · Vector Algebra · medium

A force vec F=hat i+2hat j+3hat k acts at the point (1,-1,2). Its moment about the origin is

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A force F⃗=i^+2j^+3k^\vec F=\hat i+2\hat j+3\hat k acts at the point (1,−1,2)(1,-1,2). Its moment about the origin is
  1. A. 7i^+j^−3k^7\hat i+\hat j-3\hat k
  2. B. −7i^+j^+3k^-7\hat i+\hat j+3\hat k
  3. C. i^−j^+6k^\hat i-\hat j+6\hat k
  4. D. −7i^−j^+3k^-7\hat i-\hat j+3\hat k
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Answer: (D) −7i^−j^+3k^-7\hat i-\hat j+3\hat k

The moment is r⃗×F⃗\vec r\times\vec F with r⃗=i^−j^+2k^\vec r=\hat i-\hat j+2\hat k: ∣i^j^k^1−12123∣=(−3−4)i^−(3−2)j^+(2+1)k^=−7i^−j^+3k^\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&2\\1&2&3\end{vmatrix}=(-3-4)\hat i-(3-2)\hat j+(2+1)\hat k=-7\hat i-\hat j+3\hat k. Reversing the order, F⃗×r⃗\vec F\times\vec r, gives the negative of this vector.
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