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Question

Show that four points whose position vectors are 6i^-7j^, 16i^-19j^-4k^, 3i^-6k^, 2i^-5j^+10k^ are coplanar.

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Solution

DISCLAIMER: Given points are not coplanar.

Let A, B, C, D be the given points. The given points will be coplanar iff any one ofthe following triads of vectors are coplanar: AB, AC, AD ; AB, BC, CD ; BC, BA, BD etc.In order to show that AB, AC, AD are coplanar, we will have to show that their scaler tripleproduct i.e. AB AC AD = 0Using, PQ =Position vector of Q - Position vector of P, we obtainNow, AB =16i^-19j^-4k^-6i^ -7j^ = 10i^-12j^-4k^ AC= 3i^-6k^-6i^ -7j^ =-3i^+7j^-6k^and, AD =2i^-5j^+10k^-6i^ -7j^ = -4i^+2j^+10k^ AB AC AD = 10-12-4-37-6-4210= 1070+12+12-30-24-4-6+28 = 84Thus, the given points are not coplanar.

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