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Question

Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0.

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Solution

We know that the distance of the point x1, y1, z1 from the plane ax+by+cz+d=0 is given byax1+by1+cz1+da2+b2+c2Distance of the point (1, 1, 1) from the plane 3x + 4y - 12z + 13 = 0The required distance =3 1 + 4 1 - 12 1 + 1332 + 42 + -122=3 + 4 - 12 + 139 + 16 + 144=813 units ... (1)Distance of the point (-3, 0, 1) from the plane 3x + 4y - 12z + 13 = 0The required distance =3 -3 + 4 0 - 12 1 + 1332 + 42 + -122=-9+0-12+139+16+144=813 units .... (2)From (1) and (2), we can say that the given points are equidistant from the given plane.

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