## 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.

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.

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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.

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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.

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.

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Complete Questions 11 and 12 from worksheet.