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Question
Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector `2hati + hatj - 2hatk`.
Solution
If `hatn` is a unit vector along the normal and p is the length of the perpendicular from origin to the plane, then the vector equation of the plane is `barr.hatn` = p.
Here, `barn = 2hati + hatj - 2hatk` and p = 42
∴ `|barn| = sqrt(2^2 + 1^2 + (-2)^2`
= `sqrt(9)`
= 3
`hatn = barn/|barn|`
= `(1)/(3)(2hati + hatj - 2hatk)`
∴ The vector equation of the required plane is
`barr.[1/3(2hati + hatj - 2hatk)]` = 42
i.e. `barr.(2hati + hatj - 2hatk)` = 126.
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