Linear Application for Vectors: Distance
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Linear Application for Vectors: Distance

[From: ] [author: ] [Date: 11-12-10] [Hit: ]
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.......
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!

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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.

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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}
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