Vector product: notes and practice questions
- The vector product (cross product) is defined only for 3D vectors, resulting in a vector perpendicular to both operands.
- It can be calculated algebraically using a determinant formula or geometrically via magnitude .
- Key properties include anti-commutativity () and that the cross product of parallel vectors is the zero vector.
- Applications include finding the area of parallelograms and triangles, determining the normal vector to a plane, calculating the volume of a parallelepiped, and finding the direction vector of the intersection of two planes.
How it is examined
Two uses only: a normal vector for a plane (AHL 3.17), and an area. Anti-commutativity means sign errors are graded, and the area of a triangle is with the half that students forget. 3 to 6 marks.
The component form of the vector product and the area of a parallelogram are given.
- The definition of the vector product of two vectors.
- Properties of the vector product.
- Geometric interpretation of .
Linking questions
- Links to other subjects: magnetic forces and fields (physics).
Practice questions
14 questions · 8 medium · 6 hardQuestion 1
MediumPaper 1 · no calculator4 marksLet and be two non-zero vectors. The vectors representing the diagonals of a parallelogram with sides and are and .
Show that .
Start by expanding the cross product . Remember the properties of the vector cross product, such as and . Then, express the magnitude of the cross product in terms of the dot product.
Question 2
HardPaper 1 · no calculator19 marksTwo spacecraft, S1 and S2, travel along straight paths, represented by the lines and respectively. The paths of the spacecraft intersect at a docking station D. A probe is located at a point P on the path of . This is shown in the following diagram.

The direction vector of is . The vector is given by , where .
The acute angle between the paths and is , where .
(a) Show that .
(b) Find the value of .
(c) Hence, find the shortest distance from the probe at P to the path .
The paths and lie on a plane, .
(d) Find a vector normal to the plane .
A satellite dish is modelled as a right circular cone with its vertex at V. The base of the cone lies in the plane and is centred at P. The path is tangent to the circular base of the cone. The volume of the cone is cubic units. The position vector of P is .
(e) Find the two possible position vectors for V.
Use the scalar product formula for the angle between two vectors, .
Square both sides of the equation from part (a) to eliminate the square root, then solve the resulting quadratic equation.
The shortest distance from a point P to a line L1 can be found using trigonometry. Consider the right-angled triangle formed by P, D, and the point on L1 closest to P. The distance is given by . Alternatively, use the vector product formula for the distance.
A normal vector to a plane containing two lines can be found by taking the vector product of their direction vectors.
The radius of the cone's base is the shortest distance from P to L1. Use the volume formula to find the cone's height, . The vertex V is located at a distance from the centre P, along the direction of the normal vector to the plane. Remember there are two possible directions along the normal.
Question 3
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 4
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 5
MediumPaper 2 · calculator6 marksA team of architects is designing a new building and needs to define a support beam's orientation. The beam must be perpendicular to two existing structural walls, Wall A and Wall B. The equations of the planes representing these walls are given by:
Wall A ():
Wall B ():
Find a Cartesian equation of the plane () that represents the orientation of the support beam, given that it passes through the origin (0, 0, 0).
Find the coordinates of the point where Wall A, Wall B, and the support beam's plane () intersect.
The normal vector of a plane is perpendicular to the plane. If a new plane is perpendicular to two other planes, its normal vector must be parallel to the cross product of the normal vectors of the two other planes. Remember that the equation of a plane is of the form , where is the normal vector.
You need to solve a system of three linear equations in three variables. You can use substitution, elimination, or a matrix method (e.g., with your GDC).
Question 6
HardPaper 1 · no calculator15 marksConsider the points given by the coordinates , , .
Find the vector .
Hence, find the exact area of triangle PQR.
Show that the Cartesian equation of the plane , which contains the triangle PQR, is .
A second plane is given by the equation . Find a vector equation for the line of intersection of the planes and .
First, find the position vectors and by subtracting the coordinates of the initial point from the terminal point. Then, compute their cross product, for example by using the determinant formula for a matrix.
The area of a triangle formed by two vectors is half the magnitude of their cross product. Use the result from part (a).
The cross product vector found in part (a) is a normal vector to the plane. Use this normal vector and the coordinates of one of the points (P, Q, or R) to determine the equation of the plane.
To find the line of intersection, you need to solve the system of equations for the two planes. You can set one variable, say , equal to a parameter . Alternatively, the direction vector of the line of intersection can be found by taking the cross product of the normal vectors of the two planes.
Question 7
MediumPaper 2 · calculator9 marksA landscape architect is designing a triangular shade sail for a patio. The vertices of the sail are defined by points A, B, and C in a 3D coordinate system, where the z-axis represents height.
The coordinates of the vertices are A, B and C, where is a positive constant representing a design parameter.
(a) Show that the vector product is given by .
(b) The architect wants to minimize the tension in the sail, which is proportional to the magnitude of the vector product of two adjacent sides. Find the smallest possible value of .
(c) Calculate the smallest possible area of the shade sail.
First, find the displacement vectors and . Then, use the formula for the cross product of two vectors.
To find the minimum magnitude, consider the square of the magnitude, . This will result in a polynomial in . You can then use calculus (finding the derivative and setting it to zero) or a GDC to find the minimum value for .
The area of a triangle formed by two vectors is half the magnitude of their cross product.
Question 8
HardPaper 2 · calculator13 marksTwo drones, Drone Alpha and Drone Beta, are flying in a 3D space. At a particular instant, Drone Alpha passes through point and then point .
(a) Find a vector equation of the line representing Drone Alpha's path.
At the same instant, Drone Beta passes through point and then point .
(b) Find a vector equation of the line representing Drone Beta's path.
(c) Hence, or otherwise, find the shortest distance between the paths of Drone Alpha and Drone Beta.
Recall that a vector equation of a line can be expressed as , where is the position vector of a point on the line and is the direction vector of the line.
Similar to part (a), identify a position vector and a direction vector for Drone Beta's path.
The shortest distance between two skew lines and is given by the formula .
Question 9
MediumPaper 1 · no calculator6 marksA modern art sculpture is in the shape of a tetrahedron with vertices at points P(2, 1, 0), Q(3, -1, 2), R(0, 2, 1), and S(4, 3, 5). The coordinates are given in metres relative to a fixed origin O.
Calculate the volume of the sculpture.
The volume of a tetrahedron with vertices A, B, C, and D can be found using the formula . First, find three vectors that share a common starting point, for example, , , and .
Question 10
HardPaper 2 · calculator21 marksConsider the non-zero vectors and . Let be the angle between and .
Using the definitions of and in terms of , and , show that .
A triangle PQR has vertices P(1, 0, 1), Q(, 2) and R(4, 1, 1), where .
The vectors and are defined as and .
It is given that and the area of triangle PQR is square units.
Find the value of .
Hence, or otherwise, find the value of .
Hence, or otherwise, find the possible values of and the corresponding values of .
Consider a new point S, the vector is defined as .
It is given that and , and the area of triangle PRS is 10 square units.
Assuming that , find the possible vectors for .
Recall the definitions of the dot product and the magnitude of the cross product in terms of the magnitudes of the vectors and the angle between them. Use the Pythagorean identity for trigonometric functions.
The area of a triangle formed by two vectors is half the magnitude of their cross product.
Use the identity from part (a) and the values you've found for the dot product and the magnitude of the cross product. Remember to calculate the magnitude of first.
Express in terms of and . Set up two equations using the given dot product and the magnitude of found in the previous part. Solve the system of equations.
If is perpendicular to both and , it must be parallel to their cross product. The area of triangle PRS can be found using the magnitude of and and the angle between them.
Question 11
MediumPaper 1 · no calculator8 marksTwo vectors are given by and , where .
(a) Find the value of for which the vectors and are orthogonal.
(b) For this value of , find the Cartesian equation of the plane that contains the vectors and and passes through the point .
Two vectors are orthogonal (perpendicular) if their scalar (dot) product is equal to zero. Set up the equation and solve for .
First, substitute the value of you found in part (a) into the expressions for and . Then, find the normal vector to the plane by calculating the vector (cross) product . Finally, use the point and the normal vector to write the equation of the plane in the form .
Question 12
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 13
MediumPaper 1 · no calculator6 marksA flat rectangular mirror is mounted on a wall. In a 3D coordinate system, with the origin at a corner of the room, the mirror lies on a plane .
One of the edges of the mirror is represented by the line with equation .
The plane also contains the point P.
Find the Cartesian equation of the plane .
To find the equation of a plane, you need a point on the plane and a vector normal (perpendicular) to the plane. You are given one point P. Can you find another point on the plane from the line equation? The direction vector of the line is parallel to the plane. How can you find a second vector parallel to the plane? The cross product of two vectors parallel to the plane will give you the normal vector.
Question 14
MediumPaper 2 · calculator6 marksA structural engineer is designing a framework and needs to define the orientation of certain surfaces. Consider two existing planar surfaces, and , with the following Cartesian equations:
Find a Cartesian equation of a third planar surface, , which is perpendicular to both and , and passes through the point .
Determine the coordinates of the point where , , and intersect.
The normal vector of a plane perpendicular to two other planes can be found using the cross product of their normal vectors. Once you have the normal vector and a point on the plane, you can determine its Cartesian equation.
You need to solve the system of three linear equations representing the three planes simultaneously. A calculator can be very helpful for this.
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