# Vectors-finding distance

• July 14th 2006, 09:30 PM
kingkaisai2
Vectors-finding distance
Find the distance of coordinate (1,0,0) from the line r=t(12i-3j-4k)
• July 14th 2006, 09:42 PM
malaygoel
Quote:

Originally Posted by kingkaisai2
Find the distance of coordinate (1,0,0) from the line r=t(12i-3j-4k)

Although there is a formula, use the following(you can derive the formula using it)
One thing I want to say is that the distance between a line and a point is the length of the perpendicular drawn from the the point to the line.
Let P(x,y,z) be a point lying on the line.(Let your point be A)
For distance AP to be minimum, line segmentAP should be perpendicular to the line.(Using this contion you could find the x,y,z).
After finding P, you could find distance AP using Distance formula.

Keep Smiling
Malay
• July 15th 2006, 07:58 PM
ThePerfectHacker
Sweetheart you are going to have to state your questions better next time you post.

Theorem: The distance from point $P$ to a line in $\mathbb{R}^3$ (3 dimensions) is,
$\frac{||\bold{v}\times \bold{QP}||}{||\bold{v} ||}$
where $\bold{v}$ is aligned with the line and $Q$ is any point on line.
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Solution
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First find $\bold{v}$ any vector aligned with your line, which for example is $\bold{v}=12\bold{i}-3\bold{j}-4\bold{k}$
Now select a point on the line, I will select for simplicity the origion $Q=(0,0,0)$. So that,
$\bold{QP}=\bold{i}$
Thus,
$\bold{v}\times \bold{QP}=\left| \begin{array}{ccc} \bold{i} & \bold{j} & \bold{k}\\ 12 & -3 & -4 \\ 1& 0&0 \end{array} \right|=-4\bold{j}+3\bold{k}$
Thus,
$||\bold{v} \times\bold{QP}||=||-4\bold{j}+3\bold{k}||=\sqrt{0^2+4^2+3^2}=5$
And,
$||\bold{v}||=\sqrt{12^2+3^2+4^2}=13$
Therefore, the distance is their quotient as in the theorem,
$\mbox{distance }=\frac{5}{13}$