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Question
Find the distance of the point `4hat"i" - 3hat"j" + hat"k"` from the plane `bar"r".(2hat"i" + 3hat"j" - 6hat"k")` = 21.
Solution
The distance of the point `"A"(bara)` from the plane `bar"r".hat"n" = p "is given by" d = | p - bar"a".hat"n"|`
Here, `bar"a" = 4hat"i" - 3hat"j" + hat"k", bar"r".(2hat"i" + 3hat"j" - 6hat"k")=21 `
`∴ barn = 2hati+3hatj-6hatk`
`∴ hatn=barn/|barn| = (2hat"i" + 3hat"j" - 6hat"k")/sqrt(2^2+3^2+(-6)^2)`
`hatn = (2hat"i" + 3hat"j" - 6hat"k")/7 `
∴ The normal form of equation of plane is
∴ `(bar"r".(2hat"i" + 3hat"j" - 6hat"k"))/7=21/7`
∴ `(bar"r".(2hat"i" + 3hat"j" - 6hat"k"))/7=3`
∴ `hatn =(bar"r".(2hat"i" + 3hat"j" - 6hat"k"))/7` & P=3
Now `bara.hatn=(4hat"i" - 3hat"j" + hat"k"). ((2hat"i" + 3hat"j" - 6hat"k"))/7`
= `(8-9+6)/7`
`bara.hatn = (-7)/7 = -1`
∴ `|bara.hatn|=1`
∴ The distance between point & plane is
`| p - bar"a".hat"n"|=|3-1|= 2`
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