A-Level Further Maths
3 min
Updated: Dec 7, 2020
The equation of a line can be represented by vectors in 3D. The line that passes through A and R can be written as:
r = a + λb
where a is the position vector of a known point on the line
b is the direction vector of the line (a vector parallel to the line)
r is the position vector of any arbitrary point on the line
λ is a scalar parameter
If you need to find the equation of a line from just two points, C and D, and you do not know the direction vector, use the fact that the vector between the two points (d - c) is the direction vector of the line:
The equation of this line is given as
r = c + λ(d-c)
If you need to find the equation of a straight three dimensional line in Cartesian form (in terms of x, y and x), convert using the vectors a and b from the vector equation r = a + λb.
The vector equation of a plane is given as:
r = a + λb + μc
where a is the position vector of a known point, A, in the plane
b and c are non-parallel, non-zero vectors in the plane
r is the position vector of any arbitrary point, R, in the plane
λ and μ are scalar parameters
The normal vector of a plane is used to describe the direction of the plane. It is the vector that is exactly perpendicular to the plane.
You can use the normal vector of a plane to write a Cartesian equation describing the plane.
Where the normal vector, n = ai + bj + ck, the Cartesian equation of the plane is given as:
ax + by + cz = d
a, b, c and d are all constants
The scalar product, a . b, is given by the magnitude of the two vectors a and b, and by the angle between them, θ:
a . b = |a| |b| cos(θ)
If θ = 90° (the two vectors are perpendicular), then the scalar product will equal zero.
If a and b are parallel, a . b = |a| |b|
If the two vectors are identical (both are a), the scalar product will equal |a|²
The equation of a plane can also be written as the scalar product of the normal vector to the plane and the position vector of any arbitrary point on the plane:
r . n = k
r is the position vector of any arbitrary point on the plane
n is the normal vector of the plane
k is a scalar constant for the plane, where k = a . n for a specific point in the plane with position vector a
The acute angle between two intersecting lines, with direction vectors a and b, is given as above.
|a| |b| represents the magnitude of each vector - use 3D Pythagoras for this.
To find the obtuse angle, subtract the acute angle from 180°.
The acute angle between the line r = a + λb and the plane r . n = k is given as above.
again, |b| |n| represents the magnitude of each vector - use 3D Pythagoras for this.
The acute angle between the plane r . n₁ = k₁ and the plane r . n₂ = k₂ is given as above.
When you have the vector equations of two lines, you can see if they intersect or not.
If the two lines are parallel, they never intersect.
Write out the two equations as column vectors
Write each row of the column vectors as simultaneous equations involving λ and μ
Solve two of the three equations for λ and μ
See if these values are consistent with the third equation
There are a number of possible outcomes:
If all three equations are consistent with the values of λ and μ, the lines intersect at those values of λ and μ
If two of the three equations have solutions, but the third does not match, the lines do not intersect.
If none of the three equations can be solved simultaneously, the lines do not intersect.
If the two lines are neither parallel nor intersect, they are skew
Unlike the above two examples, there is a fixed equation that can be used to find the perpendicular distance between a point and a plane:
All you have to do is substitute in the values.