Find the distance from the point (2, 3, 2) to the line x=0, y=3+3t, z=2+1t.
please show your work. best answer 10 points! Thanks!
please show your work. best answer 10 points! Thanks!
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A vector along the line is <0, 3, 1>
A vector from the point to the line is <-2, 3t, t>
The dot product of those is zero
0 + 9t + t = 0
The vectors are orthogonal when t = 0
The point on the line is (0, 3, 2)
The distance is sqrt(4 + 0 + 0) = 2
I hope I didn't put the wrong numbers in at any point - pls check carefully.
A vector from the point to the line is <-2, 3t, t>
The dot product of those is zero
0 + 9t + t = 0
The vectors are orthogonal when t = 0
The point on the line is (0, 3, 2)
The distance is sqrt(4 + 0 + 0) = 2
I hope I didn't put the wrong numbers in at any point - pls check carefully.
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The distance formula:
D = sqrt[(x - x0)^2 + (y - y0)^2 + (z - z0)^2]
Where V = (0,3+3t, 2+1t) and V0 = (2,3,2).
The order of the subtraction can be changed with no consequences because directions must be real values and they are squared which forces them to positive.
D = sqrt[(-2)^2 + (3t)^2 + (1t)^2]
= sqrt[4 + 10t^2}
D = sqrt[(x - x0)^2 + (y - y0)^2 + (z - z0)^2]
Where V = (0,3+3t, 2+1t) and V0 = (2,3,2).
The order of the subtraction can be changed with no consequences because directions must be real values and they are squared which forces them to positive.
D = sqrt[(-2)^2 + (3t)^2 + (1t)^2]
= sqrt[4 + 10t^2}