# Lines

## Lines

Let’s image we have a point in space Point(A) -2 ; 1 ; 4 and Point(B) 10 ; -7 ; 13 we can define Vector(N) for this line:

\overrightarrow{AB} \\
= Bx - Ax ; By - Ay ; Bz - Az \\
= 10 - (-2) ; (-7) - 1 ; 13 - 4 \\
\overrightarrow{AB}= 12 ; -8 ; 9 \\
I = 12 ; J = -8 ; K = 9

Now we have the vector of the line we can determine any point along the vector by using one of the points as a reference, by translating the reference point along the vector by λ. (Note:- the Vector must be Unit Vector form).

The vector equation of the line is as follows:

\utilde{R} = \begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}
+ \lambda
\begin{pmatrix}
12 \\
-8 \\
9
\end{pmatrix}

Lets find a point 25units from the reference point of the line:

### Vector Magnitude

||N|| = \sqrt{I^2 + J^2 + K^2} \\
= \sqrt{12^2 + -8^2 + 9^2} \\
= \sqrt{ 144 + 64 + 81} \\
= \sqrt{ 289 } \\
= 17

### Unit Vector of Vector(N)

\widehat{N} = \overrightarrow{N} / ||{N}|| \\
= 12 / 17 ; -8 / 17 ; 9 / 17 \\
= 0.7058 ; -0.4705 ; 0.5294

Now lets do the calculation moving the point 25units along the line in the direction of the lines vector:

\utilde{R} = \begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}
+ 25
\begin{pmatrix}
0.7058 \\
-0.4705 \\
0.5294
\end{pmatrix}\\
\utilde{R} =\begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}
+ (17.645 ; -11.7625 ; 13.235 ) \\
\utilde{R} = 15.645 ; -10.7625 ; 17.235