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Question
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.
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Solution
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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