Coincident, Parallel, intersecting, skew lines: notes and practice questions
- Lines in 2D can be coincident, parallel, or intersecting based on their slopes and y-intercepts.
- In 3D, lines are classified as parallel, coincident, intersecting, or skew.
- Parallelism in 3D is checked by comparing direction vectors for scalar multiples.
- If not parallel, lines are intersecting if a common point exists (consistent parametric equations) or skew if they don't intersect and aren't parallel.
- The angle between lines uses the dot product of direction vectors; a zero dot product indicates perpendicularity.
- The shortest distance between skew lines involves the cross product of direction vectors and the vector connecting points on each line.
How it is examined
The method is fixed: check whether the direction vectors are parallel, then solve two of the three component equations and test the third. Showing that the third equation fails is what earns the "skew" mark, and a student who writes "skew" without that check gets the word and not the mark. 5 to 7 marks, Paper 1.
- Coincident, parallel, intersecting and skew lines, distinguishing between these cases.
- Points of intersection.
Linking questions
- The three-dimensional analogue of AHL 1.16, where the same three outcomes appear algebraically.
Practice questions
8 questions · 1 easy · 4 medium · 3 hardQuestion 1
EasyPaper 1 · no calculator6 marksThe path of a drone, , can be modelled by the vector equation . A communication beacon is located at point B with coordinates . The drone passes through the location of the beacon.
(a) Find the value of .
A second drone, , starts at the point and travels on a path parallel to .
(b) Write down a vector equation for the path of .
A point lies on a line if its coordinates satisfy the line's equation for some value of the parameter. Set up three separate equations for the x, y, and z coordinates and solve for the parameter first using the known coordinates.
Parallel lines share the same direction vector. What is the direction vector of drone ? What is the position vector for the starting point of drone ?
Question 2
MediumPaper 1 · no calculator8 marksThe paths of two submarines, A and B, are described by the vector equations below, where are time parameters in hours, and the coordinates are in kilometres.
(a) Show that the paths of the two submarines do not cross.
(b) Find the shortest distance between the paths of the two submarines.
To show the paths don't cross, you need to demonstrate that the system of linear equations formed by equating the position vectors has no solution. Set the x, y, and z components equal to each other and try to solve for the parameters and .
The paths are skew lines. The shortest distance between two skew lines can be found using a formula involving the scalar triple product. Alternatively, you can define a vector between a general point on each line and use calculus or vector properties to minimize its length.
Question 3
HardPaper 1 · no calculator19 marksConsider the line with vector equation .
(a) (i) Find a Cartesian equation for the line .
(ii) Show that the point lies on .
Consider a second line with direction vector . The acute angle between and is , where .
(b) Find the possible values of .
Let a third line, , be defined by the vector equation . The lines and intersect at a point B.
(c) Find the value of and the coordinates of B.
Express the parameter in terms of , , and from the components of the vector equation. Then, set these expressions for equal to each other.
Substitute the coordinates of point P into the Cartesian equation you found in part (a)(i) and verify that the equalities hold. Alternatively, use the vector equation and show that there is a single value of that produces the point P.
Recall the formula for the angle between two vectors using the scalar (dot) product: . Since the angle between lines is typically taken as the acute angle, use the formula . This will lead to a quadratic equation in .
For the lines to intersect, there must be values of and that make their vector equations equal. Set the corresponding and components equal to each other to form a system of three linear equations in three variables (). Solve for and using two of the equations, then substitute into the third to find . Finally, use either or in its respective line equation to find the coordinates of the intersection point.
Question 4
MediumPaper 1 · no calculator6 marksThe line has vector equation , where .
The line has vector equation , where .
The lines and are perpendicular and intersect at a single point.
Find the value of and the value of .
For two lines to be perpendicular, what must be true about their direction vectors? For two lines to intersect, what must be true about their position vectors at the point of intersection?
Question 5
HardPaper 1 · no calculator19 marks(a) The line passes through the point Q(2, 0, 5) and has a direction vector .
Write down a vector equation for .
(b) A second line, , passes through the points C(3, 1, 0) and D(4, 3, -2).
Find a vector equation for .
(c) Show that and are skew.
(d) Find in terms of , where M is a general point on .
(e) Hence, find the coordinates of the point M on that is closest to Q.
(f) The origin is denoted by O(0, 0, 0). Find the equation of the plane that contains the points O, Q and the point M found in part (e). Give your answer in the form , where .
The vector equation of a line is given by , where is the position vector of a point on the line and is the direction vector of the line.
To find the vector equation of a line passing through two points, first find the direction vector by subtracting the position vectors of the two points. Then use one of the points as the position vector in the equation.
To show that two lines are skew, you must demonstrate two things: they are not parallel, and they do not intersect. Check if their direction vectors are scalar multiples of each other. Then, set the vector equations equal to each other and try to solve the resulting system of linear equations.
First, express the position vector of a general point M on using the parameter . Then find the vector by subtracting the position vector of Q from the position vector of M. Finally, calculate the scalar (dot) product of and the direction vector of , which is .
The point M on closest to Q is such that the vector is perpendicular to the direction vector of . This means their scalar product is zero. Use your result from part (d).
To find the equation of a plane, you need a point on the plane and a normal vector. You have three points (O, Q, M). You can form two vectors in the plane, for example and . The normal vector to the plane is perpendicular to both of these vectors, so you can find it by calculating their vector (cross) product.
Question 6
MediumPaper 1 · no calculator12 marksPoints P and Q have position vectors and respectively, relative to an origin O. Let M be the midpoint of the line segment [PQ].
Show that the position vector of M is .
A triangle has vertices P(1, 0, 2), Q(3, 4, -2), and R(5, 2, 6).
Let L, M and N be the midpoints of the sides [PQ], [QR] and [RP] respectively. Find the position vectors of L, M and N.
The centroid G of the triangle PQR has position vector .
Find the coordinates of G.
Show that the points P, G, and M are collinear, where M is the midpoint of [QR].
Hence, find the ratio PG:GM.
Consider the vector path from O to M. You can go directly, or via P. How can you express the vector in terms of the vector ?
Recall the midpoint formula that you just proved in part (a). Apply it to the position vectors of the vertices of the triangle.
The position vectors , , and are given by the coordinates of the points P, Q, and R. Substitute these into the given formula for the centroid.
To show three points are collinear, you can show that the vector connecting the first two points is a scalar multiple of the vector connecting the first and third points. For example, show that for some scalar .
The relationship between the vectors you found in part (d), and , directly tells you the ratio. If , what does this mean about the position of G on the line segment PM?
Question 7
HardPaper 1 · no calculator14 marksLet P(1, 0, 1), Q(1, 2, 0), and R(k+1, 1, -1) be three points in , where k > 0.
Let be the plane containing the points P, Q, and R.
(a) Find a Cartesian equation for the plane in terms of k.
(b) Let N be the midpoint of the line segment [PR]. A line L passes through N and is perpendicular to the plane . Find a vector equation for the line L in terms of k.
(c) Let be the line defined by the equations . Show that the line L does not intersect the line for any k > 0.
To find the equation of a plane, you need a point on the plane and a vector normal to the plane. You can find the normal vector by taking the cross product of two non-parallel vectors that lie in the plane, such as and .
The direction vector of a line perpendicular to a plane is the normal vector of that plane. You also need a point on the line, which is given as the midpoint of [PR].
To check for intersection, set the corresponding components of the two lines' equations equal to each other. This will give you a system of equations. Try to solve this system and see if you arrive at a contradiction.
Question 8
MediumPaper 1 · no calculator8 marksThe flight paths of two small drones, A and B, are modelled by the vector equations:
where and are real parameters.
A robotics engineer needs to determine if their paths will cross. Determine if the flight paths are skew.
To determine if two lines are skew, you first need to check if they are parallel. If they are not parallel, you then need to check if they intersect. If they are not parallel and do not intersect, they are skew.
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