AcademyVectors In Rn
Academy
Lines In R3
Level 1 - Linear Algebra topic page in Vectors In Rn.
Principle
A line in \(\mathbb R^3\) is determined by one point and one non-zero direction vector. The line consists of all points reached by starting at the fixed point and moving any real multiple of the direction vector.
Notation
\(\mathbf r\)
a variable position vector for a point on the line
\(\mathbf r_0\)
a fixed position vector for one known point on the line
\(\mathbf d\)
a non-zero direction vector for the line
\(t\)
a real parameter
\((x,y,z)\)
coordinates of a variable point on the line
The Core Method
Vector form of a line is
Vector equation of a line
\[\mathbf r=\mathbf r_0+t\mathbf d\]
If \(\mathbf r_0=(x_0,y_0,z_0)\) and \(\mathbf d=(a,b,c)\), then
Parametric line
\[(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)\]
which means
Coordinate equations
\[x=x_0+at,\qquad y=y_0+bt,\qquad z=z_0+ct\]
To test whether a point lies on the line, solve the three coordinate equations for the same value of \(t\). One shared value means the point lies on the line; inconsistent values mean it does not.
Worked Cases
Question
Find a vector equation of the line through \(P=(1,2,-1)\) and \(Q=(4,0,5)\).
Answer
Use \(P\) as the fixed point: \(\mathbf r_0=(1,2,-1)\). A direction vector is \(\overrightarrow{PQ}=Q-P=(4-1,0-2,5-(-1))=(3,-2,6)\). Therefore the line is \(\mathbf r=(1,2,-1)+t(3,-2,6)\), where \(t\in\mathbb R\).
Examples
Question
Do \(\mathbf r=(0,1,2)+t(2,4,6)\) and \(\mathbf r=(0,1,2)+s(1,2,3)\) describe the same line?
Answer
Yes. The fixed point is the same, and \((2,4,6)=2(1,2,3)\). Scaling a non-zero direction vector does not change the set of points on the line; it only changes the speed at which the parameter moves along the line.