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Home » 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.
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.
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Three Dimensional Geometry
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.

According to ques,

Coordinates are

    \[A\text{ }\left( 2,\text{ }3,\text{ }4 \right)\text{ }and\text{ }B\text{ }\left( 4,\text{ }5,\text{ }8 \right)\]

Again,

Coordinates of the mid-point C are

    \[\left( 2+4/2,\text{ }3+5/2,\text{ }4+8/2 \right)\text{ }=\text{ }\left( 3,\text{ }4,\text{ }6 \right)\]

Since the direction ratios of the normal to the plane = direction ratios of AB,

Hence,

    \[=\text{ }4\text{ }\text{ }2,\text{ }5\text{ }\text{ }3,\text{ }8\text{ }\text{ }4\text{ }=\text{ }\left( 2,\text{ }2,\text{ }4 \right)\]

Therefore,

Equation of the plane :

    \[a\left( x\text{ }\text{ }{{x}_{1}} \right)\text{ }+\text{ }b\left( y\text{ }\text{ }{{y}_{1}} \right)\text{ }+\text{ }c\left( z\text{ }\text{ }{{z}_{1}} \right)\text{ }=\text{ }0\]

Or,

    \[2\left( x\text{ }\text{ }3 \right)\text{ }+\text{ }2\left( y\text{ }\text{ }4 \right)\text{ }+\text{ }4\left( z\text{ }\text{ }6 \right)\text{ }=\text{ }0\]

Or,

    \[2x\text{ }\text{ }6\text{ }+\text{ }2y\text{ }\text{ }8\text{ }+\text{ }4z\text{ }\text{ }24\text{ }=\text{ }0\]

Or,

    \[2x\text{ }+\text{ }2y\text{ }+\text{ }4z\text{ }=\text{ }38\]

Or,

    \[x\text{ }+\text{ }y\text{ }+\text{ }2z\text{ }=\text{ }19\]

 

Required equation of plane :

    \[x\text{ }+\text{ }y\text{ }+\text{ }2z\text{ }=\text{ }19\]

or
NCERT Exemplar Solutions Class 12 Mathematics Chapter 11 - 13

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