- #1

chwala

Gold Member

- 2,745

- 387

- 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##

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: