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Question
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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Solution
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 = 1
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