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.

Answer 1

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

Coordinates of mid-point C are `((2 + 4)/2, (3 + 5)/2, (4 + 8)/2)` = (3, 4, 6)

Now 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

Hence, the required equation of plane is 

x + y + 2z = 19 or `vec"r"(hat"i" + hat"j" + 2hat"k")` = 19.