Find the shortest distance between the given vectors in 3D

  • #1

chwala

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Homework Statement
Find the shortest distance from ##(6,-4,4)## to the line joining ##(2,1,2)## and ##(3,-1,4)##
Relevant Equations
vectors in 3D
This is a textbook problem...the only solution given is ##3.##...with no working shown or given.

My working is below; i just researched for a method on google, i need to read more in this area...use of the directional vector may seem to be a more solid approach.

Ok i let ##A=(6,-4,4)##, ##B=(2,1,2)## and ##C=(3,-1,4)##
The shortest distance will be given by the formula;

##\dfrac {|BA×BC|}{|BC|}##
where
##BA=4i-5j+2k##
##BC=i-2j+2k##
therefore on substituting into the formula we shall have,
##\dfrac {|-6i-6j-3k|}{|i-2j+2k|}##= ##\dfrac {\sqrt{36+36+9}}{\sqrt {1+4+4}}=\dfrac{9}{3}=3##
 
Last edited:
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  • #2
I just realized that also
##\dfrac {|CA×CB|}{|CB|}## works! and same result is found.
 

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