Unit 3: Lines and Planes
3.1 Lines in R2
Vector Equation:
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[x, y] = [x1y1] + t [xm, ym], where:
o [x, y] represents any point on the line
o P (x1y1) represents a fixed point on the line
o t is the parameter
o m = [xm, ym] is the direction vector of the line
Points on the line can be found by choosing values for t and substituting it into the
equation
Parametric Equations:
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By rearranging the vector equation, the parametric equations can be found
x = x1 + txm
y = y1 + tym
Symmetric Equations:
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Rearranging the parametric equations and solving for t gives:
(x - x1)/xm = (y - y1)/ym
The symmetric equations represent a ratio.
3.2 Lines in R3
Vector Equation:
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[x, y, z] = [x1, y1, z1] + t [xm, ym, zm]
Parametric Equations:
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x = x1 + txm
y = y1 + tym
z = z1 + tzm
Symmetric Equations:
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(x - x1)/xm = (y - y1)/ym = (z - z1)/zm
Intersections of Lines in R³:
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Case 1: The lines intersect at one point
o The lines will have direction vectors that are NOT
scalar multiples of each other
o Find the parametric equations of each line
o Set the x, y, and z values equal to each other to create
a system of three equations and two unknowns (the
two parameters)
o Solve for the two parameters using the first two
equations and do a left side - right side check in the
third equation
o If LS = RS, the lines intersect at one point
o To find the point of intersection, sub the parameter
values into the parametric equations to find the x, y,
z coordinates
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Case 2: The lines are parallel
o The two lines have direction vectors that are scalar
multiples of each other
o There are 2 sub-cases:
 Parallel and distinct
 The lines are different and will never
intersect
 Parallel and coincident
 They are the same line and have an
infinite number of intersections
 To find which sub-case it is:
 Find the symmetric equations of each line
 Take the fixed point from line one and substitute it into the ratios of
the symmetric equations of line two
 Simplify the ratios
 If the ratios give different values, the lines are distinct
 If the ratios give the same value, the lines are coincident
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Case 3: Skew
o The two lines are not parallel but they do not
intersect
o Repeat the steps with Case 1
o LS will not equal RS
3.3 Planes
A plane is uniquely determined if you know a point and two non-collinear vectors on the plane
(basis vectors).
A plane is commonly represented with Π (pi) notation.
Vector Equation:
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[x, y, z] = [x1, y1, z1] + t [xu, yu, zu] + k [xv, yv, zv], where:
[x, y, z] represents any point on the plane
P (x1, y1, z1) represents a fixed point on the plane
t and k are the parameters
u = [xu, yu, zu] and v = [xv, yv, zv] are the basis vectors of the plane
Parametric Equations:
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x = x1 + txu + kxv
y = y1 + tyu + kyv
z = z1 + tzu + kzv
Scalar Equation (Also known as Cartesian Equation):
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Planes cannot be represented with symmetric equations, so we use scalar equations instead
The scalar equation is the most useful way of expressing planes as each plane has only one
scalar equation (whereas a plane can be represented by many different vector and
parametric equations)
A plane is uniquely defined by its normal vector, n
Ax + By + Cz + D = 0, where
o n = [A, B, C]
The normal to a plane can be found by find the cross product of any two vectors on the
plane
o n=uxv
o To find D, sub in the x, y, z coordinates of any point on the plane and solve for D
To find if a given point lies on a plane, sub in its coordinates and verify that the left side
equals 0
3.4 Problems with Lines and Planes
There are 3 possible intersections for a line and a plane.
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Case 1: Intersect at one point
o The system is consistent independent
o Change the line into parametric equations
o Sub the parametric equations into the scalar equation
of the plane
o Solve for t
o The solution will be a real number
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Case 2: Line and plane are coincident
o The system is consistent dependent
o Repeat steps for case 1
o You will end with 0t = 0
o There is an infinite number of solutions; therefore, the
solution is the equation of the line itself
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Case 3: Line and plane are parallel
o The system is inconsistent
o Repeat steps for case 1
o You will end with 0t = any number
o There are no solutions
To find the angle between a line and a plane:
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Find the angle between the line's direction vector and the plane's normal using the dot
product formula
Subtract your answer from 90°
3.5 Problems with Two Planes
There are 3 possible intersections for 2 planes.
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Case 1: Planes intersect in a line
o The intersection is a line common to both planes
o The solution is the parametric equations of the line
Case 2: Planes are parallel
o The system is inconsistent
o The planes have the same normal
o There is no solution
Case 3: Planes are coincident
o The system is consistent dependent
o The two equations represent the same plane
o There is an infinite number of solutions
Example:
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Find the intersection of the following planes: 3x - y + 4z - 7 = 0 and x + y - 2z + 5 = 0.
o Set up two system of equations and for each, eliminate a different variable
o System 1: Eliminate y
 3x - y + 4z - 7 = 0
 x + y - 2z + 5 = 0 → Add
 4x + 2z +2 = 0
 Isolate z:
 z = - 1 - 2x
o System 2: Eliminate z
 3x - y + 4z - 7 = 0
 2x + 2y - 4z + 10 = 0 → Multiply plane 2 by 2 and add
 5x + y + 3 = 0
 Isolate y:
 y = -3 - 5x
o Introduce a parameter
 Let x = t
o The parametric equations are:
 x=t
 y = -3 - 5t
 z = -1 - 2t
The direction vector of the line of intersection is the cross product of the two planes' normals.
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m = n 1 x n2
The angle between two planes is the angle between their normals.
Two planes are perpendicular if their normals dot to 0.
Linear Combinations of Planes:
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If A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0 are two planes that intersect in a
line, then a third plane containing the line of intersection can be written as a linear
combination of plane 1 and 2.
The equation of the third plane can be written as:
o A1x + B1y + C1z + D1 + k (A2x + B2y + C2z + D2) = 0
o Where k is a real number
3.6 Problems with Three Planes
There are 5 cases when dealing with 3 planes.
Case 1: All 3 planes are parallel.
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This means all 3 normal are collinear and coplanar (scalar multiples of each other).
Case 2: 2 planes are parallel and the third intersects both at the same angle.
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This means 2 normal are collinear and coplanar, and the 3rd is distinct.
Case 3: The planes intersect in pairs of lines.
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There are no collinear normals but all are coplanar.
This means each normal can be written as a linear combination of the other two.
You will see something that resembles a triangular prism.
E.g. if the three planes are:
1. x - 3y + 2z + 7 = 0
n1 = [1, -3, 2]
2. 4x + y - z + 5 = 0
n2 = [4, 1, -1]
3. 6x - 5y + 3z -1 = 0
n3 = [6, -5, 3]
 By inspection, we can see that
o n3 = 2n1 + n2
Since one normal can be written as a linear combination of the other two, the planes
intersect in pairs of lines.
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Case 4: The planes intersect in one line.
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There are no collinear normals but all are coplanar.
One plane can be written as a linear combination of the other two.
E.g. if the three planes are:
1. x - 3y + 2z + 7 = 0
n1 = [1, -3, 2]
2. 4x -5y - z + 5 = 0
n2 = [4, -5, -1]
3. 6x - 11y + 3z + 19 = 0
n3 = [6, -11, 3]
 By inspection, we can see that
o n3 = 2n1 + n2
However, we can also see that the D values agree with this statement. Therefore,
o Π3 = 2Π1 + Π2
Since one plane can be written as a linear combination of the other two, the planes
intersect in one line.
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Case 5: The planes intersect at a single point.
o
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The normals are not coplanar and will produce a unique solution.
To test for this case, use the Triple Scalar Product. If the solution is not zero, then
the planes intersect at one point.
E.g. if the three planes are:
1. 3x - 3y - 2z -14 = 0
n1 = [3, -3, -2]
2. 5x + y - 6z -10 = 0
n2 = [5, 1, -6]
3. x - 2y + 4z - 9 = 0
n3 = [1, -2, 4]
o Triple Scalar Product:
 n1 ⋅ (n2 x n3)
 = [3, -3, -3] ⋅ [4-12, -6-20, -10-1]
 = [3, -3, -3] ⋅ [-8, -26, -11]
 = 76
 ≠0
Since the solution to the Triple Scalar Product is not zero, the planes intersect at a single
o
point.
You may be asked to find the point of intersection. There are 2 ways to do this:
1. Solve by substitution (set up a system of 3 equations and solve for x, y,
and z).
2. Solve using matrix (this is usually not required for grade 12 but some
people find it faster).