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Question
Reduce the equation `bar"r".(3hat"i" + 4hat"j" + 12hat"k")` to normal form and hence find
(i) the length of the perpendicular from the origin to the plane
(ii) direction cosines of the normal.
Solution
The normal form of equation of a plane is `bar"r".hat"n" = p` where `hat"n"` is unit vector along the normal and p is the length of perpendicular drawn from origin to the plane.
Given pane is `bar"r".(3hat"i" + 4hat"j" + 12hat"k")` = 78 ...(1)
`bar"n" = 3hat"i" + 4hat"j" + 12hat"k"` is normal to the plane
∴ `|bar"n"| = sqrt(3^2 + 4^2 + 12^2) = sqrt(169)` = 13
Dividing both sides of (1) by 13, get
`bar"r".((3hat"i" + 4hat"j" + 12hat"k")/13) = (78)/(13)`
i.e. `bar"r".(3/13hat"i" + 4/13hat"j" + 12/13hat"k")` = 6
This is the normal form of the equation of plane.
Comparing with `bar"r".hat"n" = p`,
(i) the length of the perpendicular from the origin to plane is 6.
(ii) direction cosines of the normal are `(3)/(13),(4)/(13),(12)/(13)`.
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