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प्रश्न
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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उत्तर
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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