Vector Equation of a Line

The equation of a line can be written in the form

\[r=a+λb\]
where \(a\) is the position vector of any point on the line and \(b\) is any vector parallel to the line.
\(b\) is referred to as the direction vector.

VIDEO INTRODUCTION


LESSON 1

Determine the vector equation of the line which is parallel to the vector \(3i+2j-4k\) and passes through the point with position vector \(i-2j+k\).

SOLUTION
State the position vector.
\[a=i-2j+k\]
State the parallel vector.
\[b=3i+2j-4k\]
Substitute vectors into the vector equation of a line
\[r=i-2j+k+λ(3i+2j-4k)\]

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Complete Question 8 from the worksheet.


LESSON 2

The line \(l\) passes through the points \(A(2,-3,1)\) and \(B(4,0,-5)\). Determine the vector equation of \(l\).
SOLUTION
Determine the position vector.
\[a=2i-3j+k\]
Determine the parallel vector
\[b=\overrightarrow{AB}\]
\[\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}\]
\[\overrightarrow{AB}=-(2i-3j+k)+(4i-5k)\]
\[\overrightarrow{AB}=2i+3j-6k\]
Substitute the vectors into the vector equation of a line
\[l=2i-3j+k+λ(2i+3j-6k)\]
NB: \(B\) could have been used as the needed position vector and \(\overrightarrow{BA}\) as the required parallel vector.

VIDEO LESSON


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Complete Question 9 from worksheet.


LESSON 3

Determine the vector equation of the line which passes through \((2,-3,1)\) and is parallel to the vector \(i-j-2k\) in
  • Vector form
  • Parametric form and
  • Cartesian form.
SOLUTION
VECTOR FORM
\[r=\begin{pmatrix} 2\\-3 \\1 \end{pmatrix}+λ\begin{pmatrix} 1\\-1 \\-2 \end{pmatrix}\]
PARAMETRIC FORM
Replace \(r\) with \(\begin{pmatrix} x\\y \\z \end{pmatrix}\) in the vector equation of the line
\[\begin{pmatrix} x\\y \\z \end{pmatrix}=\begin{pmatrix} 2\\-3 \\1 \end{pmatrix}+λ\begin{pmatrix} 1\\-1 \\-2 \end{pmatrix}\]
Write expressions for \(x, y\) and \(z\) in terms of \(λ\)
\(x=2+λ\:\:\:\:\:\:(1)\)
\(y=-3-λ\:\:\:\:\:(2)\)
\(z=1-2λ\:\:\:\:\:\:(3)\)
These three equations are the parametric form of the equation of this particular line in terms of \(λ\).

CARTESIAN FORM

Eliminate the parameter, \(λ\)
From
(1): \(x-2=λ\)
(2): \(-3-y=λ\)
(3): \({z-1\over -2}=λ\)
Since \(λ=λ=λ\)
\(x-2=-3-y={z-1\over -2}\)
\(⟹-2x+4=6+2y=z-1\)
The Cartesian Form must be independent of the parameter.

VIDEO LESSON


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Complete Question 10 from worksheet.


LESSON 4

Show that the following pair of lines is parallel.
\[L:5i+3j+4k+λ(-i+2j+3k)\]
\[N:r=4i-2j+k+μ(3i-6j-9k)\]
SOLUTION
We simply need to show that the two direction vectors are parallel.
Since \(3i-6j-9k=3(-i+2j+3k)\) the two lines are parallel.

VIDEO LESSON


LESSON 5

Show that the following pair of lines intersect and determine the point of intersection.

\[L:r=4i-3j+k+λ(i+2j-k)\]
\[N:r=2i+6j-k+μ(-5i+3j+k)\]
SOLUTION
Since the line intersect there must exist \(λ\) and \(μ\) such that
\[4i-3j+k+λ(i+2j-k)=2i+6j-k+μ(-5i+3j+k)\]
\[(4+λ)i+(-3+2λ)j+(1-λ)k=(2-5μ)i+(6+3μ)j+(-1+μ)k\]
Equate coefficients of \(i\)
\(4+λ=2-5μ\)
\(λ+5μ=-2\:\:\:\:\:(1)\)
Equate coefficients of \(j\):
\(-3+2λ=6+3μ\)
\(2λ-3μ=9\:\:\:\:\:(2)\)
Solve (1) and (2)
\(μ=-1\)
\(λ=3\)
We will now equate the coefficients of \(k\) to determine if the values of \(λ\) and \(μ\) are consistent.
Equate coefficients of \(k\):
\(1-λ=-1+μ\)
\(1-3=-1+(-1)\)
\(-2=-2\)
The values are consistent therefore \(L\) and \(N\) intersect.

We simply substitute \(λ=3\) into \(L\) or \(μ=-1\) into \(N\) to determine the point of intersection.
Using \(λ=3\)
\(4i-3j+k+3(i+2j-k)=7i+3j-2k\)
Point is \(\begin{pmatrix} 7\\3 \\-2 \end{pmatrix}\)
NB: If the values for \(λ\) and \(μ\) are inconsistent and the lines are not parallel they are referred to as SKEWED.

VIDEO LESSON


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Complete Questions 11 and 12 from worksheet.