The distance of the point 2hati + hatj - hatk from the plane vecr.(hati - 2hatj + 4hatk) = 9 will be ______.

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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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