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Question
A plane passes through the point (− 1, 1, 2) and the normal to the plane of magnitude `3sqrt(3)` makes equal acute angles with the coordinate axes. Find the equation of the plane
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Solution
Given magnitude = `3sqrt(3)` and `vec"a" = -vec"i" + vec"j" + 2vec"k"`
Then, the normal vector makes equal acute angle with the coordinate axes.
We know that cos2 α + cos2 β + cos2 γ = 1
But α = β = γ
`3cos^2alpha` = 1
⇒ `cos^2alpha = 1/3`
`cosalpha = 1/sqrt(3)`
∴ So `vec"n" = 3sqrt(3) [1/sqrt(3)vec"i" + 1/sqrt(3)vec"j" + 1/sqrt(3)vec"k"]`
`vec"n" = 3vec"i" + 3vec"j" + 3vec"k"`
Vector equation of the plane
`vec"r"*vec"n" = vec"a"*vec"n"`
`vec"r"*vec"n" = (-vec"i" + vec"j" + 2vec"k")(3vec"i" + vec"j" + 3vec"k")`
`vec"r"*(hat"i" + 3hat"j" + 3hat"k") = -3 + 3 + 6`
`vec"r"*(hat"i" + 3hat"j" + 3hat"k")` = 6
or
`vec"r"*(vec"i" + vec"j" + vec"k")` = 2
Cartesian equation of the plane
`(xvec"i" + yvec"j" + zvec"k")*(3vec"i" + 3vec"j" + 3vec"k)` = 6
⇒ 3x + 3y + 3z = 6
or
⇒ x + y + z = 2
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