## 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.