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Question

Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).

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Solution



The required plane passes through the point P (2, 5, -3) whose position vector is a=2 i^+5 j^-3 k^ and is normal to the vector n given byn = PQ×PR.Clearly, PQ = OQ - OP = -2 i^ - 3 j^ + 5 k^ - 2 i^ + 5 j^ - 3 k^ = -4 i ^- 8 j^ + 8 k^PR = OR - OP = 5 i^ + 3 j^ - 3 k^ - 2 i^ + 5 j ^- 3 k^ = 3 i ^- 2 j^ - 0 k^n = PQ × PR = i^j^k^-4-883-20 = 16 i^ + 24 j^ + 32 k^The vector equation of the required plane isr. n = a. nr. 16 i^+24 j^+32 k^ = 2 i^+5 j^-3 k^. 16 i^+24 j^+32 k^r. 8 2 i^+3 j^+4 k^ = 32+120-96r. 8 2 i^+3 j^+4 k^ = 56r. 2 i^+3 j^+4 k^ = 7

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