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Solve the following : Find the vector equation of the plane which bisects the segment joining A(2, 3, 6) and B(4, 3, –2) at right angle. - Mathematics and Statistics

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प्रश्न

Solve the following :

Find the vector equation of the plane which bisects the segment joining A(2, 3, 6) and B(4, 3, –2) at right angle.

योग

उत्तर

The vector equation of the plane passing through `"A"(bara)` and perpendicular to the vector `bar"n"` is `bar"r".bar"n" = bar"a".bar"n"`            ...(1)

The position vectors `bar"a" and bar"b"` of the given points A and B are `bar"a" = 2hat"i" + 3hat"j" + 6hat"k" and bar"b" = 4hat"i" + 3hat"j" - 2hat"k"`

If M is the midpoint of segment AB, the position vector `bar"m"` of M is given by

`bar"m" = (bar"a" + bar"b")/(2)`

= `((2hat"i" + 3hat"j" + 6hat"k") + (4hat"i" + 3hat"j" - 2hat"k"))/(2)`

= `(6hat"i" + 6hat"j" + 4hat"k")/(2)`

= `3hat"i" + 3hat"j" + 2hat"k"`

The plane passes through `"M"(bar"m")`.

AB is perpendicular to the plane

If `bar"n"` is normal to the plane, then `bar"n" = bar"AB"`

∴ `bar"n" = bar"b" - bar"a" = (4hat"i" + 3hat"j" - 2hat"k") - (2hat"i" + 3hat"j" + 6hat"k")`

= `2hat"i" - 8hat"k"`

∴ `bar"m".bar"n" = (3hat"i" + 3hat"j" + 2hat"k").(2hat"i" - 8hat"k")`

= (3)(2) + (3)(0) + (2)(– 8)
= 6 + 0 – 16
= – 10
∴ from (1), the vector equation of the required plane is `bar"r".bar"n" = bar"m".bar"n"`

i.e. `bar"r".(2hat"i" - 8hat"k")` = – 10

i.e. `bar"r".(hat"i" - 4hat"k")` = – 5.

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Vector and Cartesian Equations of a Line
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Line and Plane - Miscellaneous Exercise 6 B [पृष्ठ २२६]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 6 Line and Plane
Miscellaneous Exercise 6 B | Q 20 | पृष्ठ २२६

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