Exam-style Questions: Vectors, Lines and Planes
1. The line L passes through the points A (3, 0, -1) and B (5, -1, 4).
a) Find the vector equation of the line L.
b) Determine whether or not the line L intersects the line with the equations
(Marks available: 6)
Answer outline and marking scheme for question: 1
Give yourself marks for mentioning any of the points below:
a) The equation for a line should be expressed as:
r = a + λb
Where a is a point on the line, b is a vector parallel to the line and λ is any number.
a = the first point A.
b = point B minus point A.
Putting these values into the equation of the line above, gives:
(2 marks)
b) Consider the point where the x values are the same for both lines, therefore:
3 + 2λ = 5 - 4μ
Consider the point where the y values are the same for both lines, therefore:
0 - 1λ = 1 + 1μ
Solving these equation simultaneously, gives:
λ= -3, μ = 2.
Putting these values into equation for line L, gives:
z = -1 + 5λ = -16
Putting these values into equation for line r, gives:
z = 11 + 3μ= 17.
As the value of z is not the same, both the line cannot be at the same point in space (i.e. they do not intersect).
(4 marks)
(Marks available: 6)
2. A body of mass 0.5 kg moves so that its velocity at time t seconds is
Find the magnitude of the momentum when t = 0 and t = 2.
(Marks available: 3)
Answer outline and marking scheme for question: 2
Give yourself marks for mentioning any of the points below:
At t = 0, the vectors equals
The magnitude of the velocity equals
Therefore the momentum at t = 0 equals
0.5 x 12.17 = 6.08 kgms-1.
Performing the same calculation at t = 2, gives the momentum equal to
0.5 x 4.47 = 2.23 kgms-1.
(Marks available: 3 marks)
3. Two lines A and B, have the following formulas:
and
a) determine whether these two lines intersect
b) find the angle between them.
(Marks available: 6)
Answer outline and marking scheme for question: 3
Give yourself marks for mentioning any of the points below:
a) Matching the x-values gives: 4 - 4λ = 6 +2μ
Matching the y-values gives: 0 + 8λ = -10 - 6μ
Matching the z-values gives: -2 -2λ = -10 -2μ
Solving the first two simultaneous equations gives: λ = 1, μ = -3.
These values work in the third equation therefore the lines meet.
Substituting λ = 1, μ = -3 into the equation of lines gives the point of intersection as being:
x = 0, y = 4, z = -2.
Therefore the lines meet at (0, 4, -2)
(3 marks)
b) The angle between the lines is the angle between the direction vectors, so using the scalar product we get,
= 
= -0.855
This gives θ = 148.8o (or 180o - 148.8o = 31.2o).
(3 marks)
(Marks available: 6)