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Class 12 Vector Algebra · Long Answer3 Marksvector_identities

Class 12 Maths · Vector Algebra · Long Answer #80

For position vectors a,b,c of triangle vertices, prove its oriented vector area is ½(b×c+c×a+a×b). Deduce the collinearity condition and unit normals.
Displacements for the oriented triangle area
Displacements for the oriented triangle area

✅ Step-by-Step Solution

Final Answer: Vector area=S/2; collinear iff S=0; unit normals ±S/|S| when S≠0, where S=b×c+c×a+a×b.
  1. The oriented vector area is ½[(b−a)×(c−a)].
  2. Expand: (b−a)×(c−a)=b×c−b×a−a×c+a×a=b×c+a×b+c×a=S.
  3. The geometric area is |S|/2. It vanishes exactly when the three points are collinear, equivalent to S=0.
  4. For a nondegenerate triangle S≠0, the two unit normals are ±S/|S|. A collinear triple has no uniquely determined triangle plane.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] The oriented vector area is ½[(b−a)×(c−a)].
  • [1 mark] Expand: (b−a)×(c−a)=b×c−b×a−a×c+a×a=b×c+a×b+c×a=S.
  • [1 mark] The geometric area is |S|/2. It vanishes exactly when the three points are collinear, equivalent to S=0.
  • [0.5 marks] For a nondegenerate triangle S≠0, the two unit normals are ±S/|S|. A collinear triple has no uniquely determined triangle plane.

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