हिंदी

Find the equation of a plane which is at a distance 33 units from origin and the normal to which is equally inclined to coordinate axis

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प्रश्न

Find the equation of a plane which is at a distance `3sqrt(3)` units from origin and the normal to which is equally inclined to coordinate axis

योग
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उत्तर

If α, β, and γ are the angles made by the line segment with the coordinate axis.
 
cosα, cosβ and cosγ are called as the direction cosines.
 
Let the required equation of the plane be `vecr * hatn` = p, where p = `3sqrt(3)`.
 
Let `vecn = (cosalpha)hati + (cosalpha)hatj + (cosalpha)hatk`, where α is acute.
 
Then, `cos^2alpha + cos^2alpha + cos^2alpha` = 1
 
⇒ `3cos^2alpha` = 1
 
⇒ `cos^2alpha = 1/3`
 
⇒ `cos alpha = 1/sqrt(3)`
 
∴ The required equation is `vecr * (1/sqrt(3)hati + 1/sqrt(3)hatj + 1/sqrt(3)hatk) = 3sqrt(3)`
 
Hence, `vecr * (hati + hatj + hatk)` = 9
 
⇒ `(xhati + yhatj + zhatk) * (hati + hatj + hatk)` = 9
 
⇒ x + y + z = 9. 
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अध्याय 12: Introduction to Three Dimensional Geometry - Exercise [पृष्ठ २३५]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 12 Introduction to Three Dimensional Geometry
Exercise | Q 8 | पृष्ठ २३५

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