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Question

Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0.

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Solution

Let Q be the image of the point P (1, 3, 4) in the plane 2x-y+z+3=0.Then PQ is normal to the plane. So, the direction ratios of PQ are proportional to 2, -1, 1.Since PQ passes through P (1, 3, 4) and has direction ratios proportional to 2, -1 and 1, equation of PQ isx-12=y-3-1=z-41=r (say)Let the coordinates of Q be 2r+1, -r+3, r+4. Let R be the mid-point of PQ. Then,R=2r+1+12, -r+3+32, r+4+42=r+1, -r+62, r+82Since R lies in the plane 2x - y + z + 3 = 0,2 r+1 - -r+62 + r+82 + 3 = 04r + 4 + r - 6 + r + 8 + 6 = 06r + 12 = 0r = -2Substituting this in the coordinates of Q, we getQ=2r+1, -r+3, r+4.=2 -2+1, 2+3, -2+4=-3, 5, 2

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