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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.
Solution
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.
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