हिंदी

Find the distance of the point whose position vector is (2i^+j^-k^) from the plane r→⋅(i^-2j^+4k^) = 9

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प्रश्न

Find the distance of the point whose position vector is `(2hati + hatj - hatk)` from the plane `vecr * (hati - 2hatj + 4hatk)` = 9

योग
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उत्तर

Here `veca = 2hati + hatj - hatk`

`vecn = hati - 2hatj + 4hatk`

And d = 9

So, the required distance is `(|(2hati + hatj - hatk)*(hati - 2hatj + 4hatk) - 9|)/sqrt(1 + 4 + 16)`

= `(|2 - 2 - 4 - 9|)/sqrt(21)`

= `13/sqrt(21)`.

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अध्याय 12: Introduction to Three Dimensional Geometry - Solved Examples [पृष्ठ २२५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 12 Introduction to Three Dimensional Geometry
Solved Examples | Q 5 | पृष्ठ २२५

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