Find the shortest distance between the given vectors in 3D

[chwala]

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,
$\dfrac {|-6i-6j-3k|}{|i-2j+2k|}$= $\dfrac {\sqrt{36+36+9}}{\sqrt {1+4+4}}=\dfrac{9}{3}=3$

 

[chwala]

I just realized that also
$\dfrac {|CA×CB|}{|CB|}$ works! and same result is found.