## 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