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in 3D Coordinate Geometry by (55.4k points)
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Find the equation of a plane which bisects perpendicularly the line joining the points A (2, 3, 4) and B (4, 5, 8) at right angles.

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Given coordinates are A (2, 3, 4) and B (4, 5, 8)

Now, the coordinates of the mid-point C are (2+4/2, 3+5/2, 4+8/2) = (3, 4, 6)

And, the direction ratios of the normal to the plane = direction ratios of AB

= 4 – 2, 5 – 3, 8 – 4 = (2, 2, 4)

Equation of the plane is

a(x – x1) + b(y – y1) + c(z – z1) = 0

2(x – 3) + 2(y – 4) + 4(z – 6) = 0

2x – 6 + 2y – 8 + 4z – 24 = 0

2x + 2y + 4z = 38

x + y + 2z = 19

Thus, the required equation of plane is x + y + 2z = 19.

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