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A plane passes through the point (− 1, 1, 2) and the normal to the plane of magnitude 33 makes equal acute angles with the coordinate axes. Find the equation of the plane - Mathematics

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प्रश्न

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

बेरीज
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उत्तर

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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Different Forms of Equation of a Plane
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Applications of Vector Algebra - Exercise 6.6 [पृष्ठ २५९]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 6 Applications of Vector Algebra
Exercise 6.6 | Q 4 | पृष्ठ २५९

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