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Question
The distance of the point `2hati + hatj - hatk` from the plane `vecr.(hati - 2hatj + 4hatk)` = 9 will be ______.
Options
13
`13/sqrt(21)`
21
`21/sqrt(13)`
Solution
The distance of the point `2hati + hatj - hatk` from the plane `vecr.(hati - 2hatj + 4hatk)` = 9 will be `underlinebb(13/sqrt(21))`.
Explanation:
Plane `vecr.(hati - 2hatj + 4hatk)` = 9 ...(1)
and Point `veca = 2hati + hatj - hatk`
Here `vecn = hati - 2hatj + 4hatk`
`|vecn| = sqrt(1^2 + (-2)^2 + 4^2)`
= `sqrt(1 + 4 + 16)`
= `sqrt(21)`
Then distance of point `veca` from plane (1)
= `(|veca.vecn - 9|)/|vecn|`
= `(|(2)(1) + (1)(-2) + (-1)(4) - 9|)/sqrt(21)`
= `(|2 - 2 - 4 - 9|)/sqrt(21)`
= `13/sqrt(21)` units.
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