Find the equation of a plane which bisects perpendicularly the line jo

1.	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.
Answer» 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.

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.

Answer»

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.