We can use projection
recall that the projection of vector u on v = (u dot v)/|v|
let the point P be such P is on the given line and OP is the length we want
A(-1,-2,3) is a point on our line and let v = <3,1,2> be the direction vector of our line
let vector AO = u = <1, 2, -3>
AP = (u dot v)/|v| = <1, 2, -3> dot <3,1,2> / |<3,1,2>|
= |(3 + 2 - 6)| /√(9+1+4) = 1/√14
now to Pythagoras:
AO = √(1+4+9) = √14 <--- hypotenuse
AP = 1/√14
OP^2 + AP^2 = AO^2
1/14 + AP^2 = 14
AP^2 = 14 - 1/14 = 195/14
AP = √(195/14) = appr 3.732 units
or
in the same triangle, find angle A
u dot v = |u| |v|cosA
-1 = √14*√14cosA
A = 94.096..
sinA = OP/AO
OP = AOsin94.096.. = 3.732 , same as before
Find the distance from the origin to the line
(x + 1)/3 = (y + 2)/1 = (z-3)/2
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