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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. - Mathematics

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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.

योग

उत्तर

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 = 1

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अध्याय 12: Introduction to Three Dimensional Geometry - Exercise [पृष्ठ २३५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 12 Introduction to Three Dimensional Geometry
Exercise | Q 7 | पृष्ठ २३५

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