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Question
Solve the following :
Reduce the equation `bar"r".(6hat"i" + 8hat"j" + 24hat"k")` = 13 normal form and hence find
(i) the length of the perpendicular from the origin to the plane.
(ii) direction cosines of the normal.
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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".(6hat"i" + 8hat"j" + 24hat"k")` = 13 ...(1)
`bar"n" = 6hat"i" + 8hat"j" + 24hat"k"` is normal to the plane
∴ `|bar"n"| = sqrt(6^2 + 8^2 +24^2) = sqrt(36 + 64 + 576) = sqrt(676) = 26`
Dividing both sides of (1) by 26, get
`bar"r".(6/26hat"i" + 8/26hat"j" + 24/26hat"k") = (13)/(26)`
`bar"r".(3/13hat"i" + 4/13hat"j" + 12/13hat"k") = 1/2`
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 `(1)/(2)`.
(ii) direction cosines of the normal are `(3)/(13),(4)/(13),(12)/(13)`.
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