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.
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 = 19 or