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Question
The sine of the angle between the straight line `(x - 2)/3 = (y - 3)/4 = (z - 4)/5` and the plane 2x – 2y + z = 5 is ______.
Options
`10/(6sqrt(5))`
`4/(5sqrt(2))`
`(2sqrt(3))/5`
`sqrt(2)/10`
Solution
The sine of the angle between the straight line `(x - 2)/3 = (y - 3)/4 = (z - 4)/5` and the plane 2x – 2y + z = 5 is `sqrt(2)/10`.
Explanation:
From equation of line we find the direction vector
`vecs` = (3, 4, 5)(l, m, n)
From equation of plane we find the normal vector
`vecq` = (2, −2, 1)(A, B, C)
Using formula,
sin θ = `|Al + Bm + Cn|/(sqrt(A^2 + B^2 + C^2 * sqrt(1^2 + m^2 + n^2)`
sin θ = `|6 - 8 + 4|/(sqrt(4 + 4 + 1) * sqrt(9 + 16 + 25)`
sin θ = `3/(3.5sqrt(2))`
sin θ = `sqrt(2)/10`
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