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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.

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

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