Class 12 Vector Algebra · Long Answer3 Marksvector_equations

Class 12 Maths · Vector Algebra · Long Answer #115

For real vectors satisfying a×b=c and a×c=b, determine what follows about mutual perpendicularity, |b|,|c| and |a|.

✅ Step-by-Step Solution

Final Answer: b=c=0; a is unrestricted, so |a|=1 does not follow.

From a×b=c, c is orthogonal to a and b. From a×c=b, b is orthogonal to a.
Take the dot product of a×c=b with b: (a×c)·b=|b|². Cyclic scalar triple products give (a×c)·b=−(a×b)·c=−|c|².
Hence |b|²=−|c|², forcing b=c=0 over the reals. Conversely, b=c=0 satisfies both equations for every a.
Thus |b|=|c|=0 and pairwise dot products vanish, but no three nonzero mutually perpendicular vectors satisfy the printed equations, and |a| is unrestricted.

[0.5 marks] From a×b=c, c is orthogonal to a and b. From a×c=b, b is orthogonal to a.
[1 mark] Take the dot product of a×c=b with b: (a×c)·b=|b|². Cyclic scalar triple products give (a×c)·b=−(a×b)·c=−|c|².
[1 mark] Hence |b|²=−|c|², forcing b=c=0 over the reals. Conversely, b=c=0 satisfies both equations for every a.
[0.5 marks] Thus |b|=|c|=0 and pairwise dot products vanish, but no three nonzero mutually perpendicular vectors satisfy the printed equations, and |a| is unrestricted.

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