Distance between point and line = [tex]\frac{\sqrt{261} }{3}[/tex]
Distance
Given that,
Given point (6, -4 , 1)
Given equation of line
x = 2t, y = t − 3, z = 2t + 2
The directional vector of the given line is
<2 , 1 , 2> at t =0 the point <0, -3 , 2> is on the line.
Distance between point and line
= | < 0-6, -3 + 4 , 2-1 > x <2 , 1, 2>| / [tex]\sqrt{2^{2} + 1^{2} + 2^{2}[/tex]
= |<-6 , 1 , 1> x <2 , 1 , 2 >| / [tex]\sqrt{9}[/tex]
<-6 , 1 , 1 > x < 2, 1, 2 > = (2 - 1)i - (-12 -2) j + (-6 -2) k
= i + 14 j - 8k
|<-6 , 1 , 1> x < 2, 1, 2>| = [tex]\sqrt{1^{2} + 14^{2} + -8^{2} }[/tex]
= [tex]\sqrt{261}[/tex]
Hence , distance between the point and line = [tex]\frac{\sqrt{261} }{3}[/tex]
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