Vector equations of a plane: notes and practice questions
- A plane can be represented by a vector equation (), parametric equations, or a scalar product form ().
- The Cartesian equation of a plane is , where is the normal vector.
- To find the equation of a plane, you can use three non-collinear points, a line and a point, or two parallel lines, often involving cross products to find the normal vector.
- Angles between lines and planes, or between two planes, are calculated using dot products involving direction vectors and normal vectors.
- Intersections can be a point (line and plane), a line (two planes), or a point, line, or no intersection (three planes).
- The distance from a point to a plane is found using a projection onto the normal vector.
How it is examined
Convert between the three forms, usually starting from three points: two displacement vectors, a cross product for the normal, then and finally the Cartesian form. The coefficients of the Cartesian equation are the normal, which is the insight the question is testing. 5 to 8 marks.
All three forms are given.
- Vector equations of a plane: , where and are non-parallel vectors within the plane.
- , where is a normal to the plane and is the position vector of a point on the plane.
- Cartesian equation of a plane .
Linking questions
- Uses AHL 3.16 to produce the normal from two vectors in the plane.
Practice questions
20 questions · 9 medium · 11 hardQuestion 1
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 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 2 · calculator18 marksTwo automated guided vehicles (AGVs), AGV-1 and AGV-2, are moving along straight paths in a 3D warehouse. The path of AGV-1 is given by the vector equation where .
The path of AGV-2 is given by the vector equation where .
All coordinates are in meters.
(a) Show that the paths of AGV-1 and AGV-2 intersect at a point P and find the position vector of P.
(b) A safety sensor plane is installed in the warehouse. The plane is given by the equation . Verify that the paths of both AGV-1 and AGV-2 lie entirely within this safety sensor plane .
(c) An emergency charging station is located at point Q with position vector .
(i) A charging drone is dispatched from Q and travels along a path perpendicular to the plane . This drone lands on the plane at point R. Find the position vector of R.
(ii) Calculate the shortest distance from the emergency charging station Q to the safety sensor plane .
(d) Due to a system malfunction, AGV-1 needs to be redirected to a virtual point Q' which is the reflection of the emergency charging station Q in the plane . Find the position vector of Q'.
To show that two lines intersect, you need to find values for the parameters (s and t) that satisfy all three component equations. Then, substitute these parameters back into one of the vector equations to find the intersection point.
For a line to lie entirely within a plane, two conditions must be met: the direction vector of the line must be perpendicular to the normal vector of the plane, and any point on the line must satisfy the plane's equation.
The line from Q to R is perpendicular to the plane, so its direction vector is the normal vector of the plane. Find the equation of this line, then find its intersection with the plane .
The shortest distance from point Q to the plane is the magnitude of the vector QR, where R is the projection of Q onto the plane.
The point R (found in part c.i) is the midpoint of the line segment QQ'. Use the midpoint formula to find the coordinates of Q'.
Question 4
HardPaper 1 · no calculator21 marksThe plane has equation .
(a) Show that the point lies on the plane .
The plane is given by , where and .
(b) In the case where , is perpendicular to and point A lies on . Given that , find the value of and the value of .
For parts (c), (d) and (e) it is now given that is parallel to .
(c) Given that , determine the value of .
It is also given that .
The line through A that is perpendicular to meets at the point B.
(d) (i) Find the coordinates of B.
(ii) Hence, find the perpendicular distance between and .
(e) Find the equation of a third parallel plane which is also a perpendicular distance of from .
To show that a point lies on a plane, substitute the coordinates of the point into the equation of the plane and verify that the equation holds true.
Recall the condition for two planes to be perpendicular in terms of their normal vectors. The dot product of the normal vectors must be zero. After finding the value of 'a', use the fact that point A lies on to find 'd'.
For two planes to be parallel, their normal vectors must be scalar multiples of each other. Set up a proportionality relationship between the components of the normal vectors.
First, write down the vector equation of the line passing through point A. The direction vector of this line is the normal vector of plane . Then, find the point of intersection of this line with plane by substituting the parametric equations of the line into the equation of the plane.
The perpendicular distance between the two parallel planes is the distance between point A (on ) and point B (on ). Calculate the magnitude of the vector .
The plane is on the opposite side of from . Point A is on . You found point B on by moving from A along the normal vector. To find a point C on , you need to move from A in the opposite direction by the same distance.
Question 5
MediumPaper 1 · no calculator4 marksConsider the two planes and defined by the equations:
where .
Show that the two planes are not perpendicular for any value of .
To determine if two planes are perpendicular, you need to consider their normal vectors. What is the relationship between the normal vectors of two perpendicular planes? Calculate the relevant quantity and see if the condition for perpendicularity can ever be met.
Question 6
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 7
MediumPaper 2 · calculator10 marksA drone's initial flight path from its base station at the origin is represented by a displacement vector . Let , , and be the angles that makes with the positive -axis, -axis, and -axis respectively.
Show that .
If the drone's initial displacement vector is meters, calculate the angles , , and to one decimal place.
A security laser beam is emitted from the base station (origin) along a direction perpendicular to the drone's initial flight path. Show that the equation of the plane containing this laser beam can be expressed in the form , where , , and are the angles found in part (b).
Consider the dot product of the vector with the unit vectors , , and . Recall that . What is the magnitude of the unit vectors?
First, calculate the magnitude of the vector . Then use the formulas for the direction cosines: , , . Remember to use the inverse cosine function to find the angles in degrees.
A plane passing through the origin has the general equation . The vector is the normal vector to the plane. How is the normal vector related to the direction of the drone's flight path?
Question 8
HardPaper 2 · calculator20 marksThree points , and lie on the plane .
Find the vector and the vector .
Hence find the equation of , expressing your answer in the form , where .
Plane has equation .
The line is the intersection of and . Verify that the vector equation of can be written as .
The plane is given by . The line and the plane intersect at the point .
Show that at the point , .
Hence find the coordinates of .
The point lies on .
Find the reflection of the point in the plane .
Hence find the vector equation of the line formed when is reflected in the plane .
To find a vector between two points, subtract the coordinates of the initial point from the coordinates of the terminal point.
The cross product of two vectors lying in a plane gives a normal vector to the plane. Then use the formula where is the normal vector and is a point on the plane.
To verify the line equation, substitute the general point of the line into the equations of both planes. Both equations should hold true for any value of . Alternatively, check if the direction vector is perpendicular to the normal vectors of both planes and if the position vector lies on both planes.
Substitute the parametric equations of line into the equation of plane and solve for .
Substitute the value of found in part (d.i) back into the vector equation of line to find the coordinates of point .
Find the equation of the line passing through and perpendicular to . Find the intersection point of this line with (this is the midpoint between and its reflection ). Use the midpoint formula to find .
The reflected line passes through point (the intersection of and ) and the reflected point found in part (e.i). Find the direction vector using these two points.
Question 9
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 10
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 11
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 12
HardPaper 1 · no calculator9 marksA laser beam is emitted from a source at point . The beam reflects off a flat mirror which lies on the plane . The reflected beam appears to originate from a virtual source at point , where is the reflection of in the plane .
Determine the coordinates of .
Find the exact distance between the laser source and the virtual source .
The line segment is perpendicular to the plane of the mirror. First, find the equation of the line that passes through and is normal to the plane. Then, find the point where this line intersects the plane. This intersection point is the midpoint of the segment .
You can use the distance formula between two points in 3D space, using the coordinates of A and the coordinates of B you found in part (a). Alternatively, you can find the perpendicular distance from point A to the plane and double it.
Question 13
MediumPaper 2 · calculator8 marksA satellite dish is positioned at a ground control station . The dish is designed to track a celestial object whose path can be modelled by a line with vector equation , where .
The plane of the satellite dish contains the line and passes through the ground control station .
Show that the Cartesian equation of the plane is .
Consider three large display screens in a museum, represented by the planes:
where .
For a special holographic effect, the three planes must intersect along a single line.
Find the value of and the value of .
To find the Cartesian equation of a plane, you need a normal vector and a point on the plane. You can find two direction vectors within the plane: one from the given line, and another by connecting a point on the line to the given point . The cross product of these two direction vectors will give you the normal vector to the plane.
For three planes to intersect in a line, the system of linear equations must have infinitely many solutions. This implies two conditions: the determinant of the coefficient matrix must be zero, and the system must be consistent (i.e., no contradictions arise during row reduction, leading to a row of zeros in the augmented matrix).
Question 14
HardPaper 2 · calculator8 marksA deep-sea probe's trajectory is modelled by a straight line with a direction vector , where is a constant. The probe needs to pass through a specific geological layer, which can be approximated by a plane with a normal vector . The efficiency of data collection is maximized when the acute angle between the probe's trajectory and the geological layer is maximized.
Determine the value of that maximizes this acute angle, and hence find the maximum acute angle. Give your answer in degrees, correct to decimal place.
Recall the formula for the angle between a line with direction vector and a plane with normal vector : . To maximize , you need to maximize . Consider maximizing to simplify differentiation.
Question 15
MediumPaper 2 · calculator11 marksConsider the three points P(4,0,1), Q(0,−3,1), and R(2,2,−5) lie on a plane
Find the vector and the vector .
Find the cartesian equation of plane .
Find the equation of the line L that passes through the point S(-8,1,23) and perpendicular to .
Find the coordinates of the point of intersection between line L and plane .
Find the difference in coordinates between the points to determine the direction vectors.
Since contains points P, Q and R so the vector product of gives the normal vector of .
A line perpendicular to the plane will have a direction vector equal to the normal vector of the plane.
Substitute the parametric expressions for x, y, and x from the line L equation into the plane equation and solve for t.
Question 16
HardPaper 1 · no calculator11 marksA plane has the Cartesian equation . A point B has coordinates .
(a) Find the vector equation of the line that passes through the point B and is perpendicular to the plane .
(b) Find the coordinates of the point of intersection, N, of the line and the plane . Hence, find the exact distance between the point B and the plane .
(c) The point P has coordinates .
Show that the distance between the point P and the plane is given by
The direction vector of a line perpendicular to a plane is the same as the normal vector of the plane. How can you find the normal vector from the plane's equation?
First, write the equation of the line in parametric form. Then, substitute these parametric equations into the equation of the plane to find the value of the parameter at the point of intersection.
You can follow the same procedure as in part (b), but use the general point instead of . Alternatively, consider the scalar projection of the vector from any point on the plane to P onto the normal vector of the plane.
Question 17
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.
Question 18
HardPaper 2 · calculator15 marksA laser beam is modelled by the line with equation . The beam strikes a flat mirror surface, which lies on the plane with equation .
(a) Find the coordinates of the point where the laser beam hits the mirror.
(b) Determine the acute angle between the laser beam and the mirror surface.
(c) Find the vector equation of the reflected laser beam.
Convert the line equation into parametric form and substitute these expressions into the plane equation to solve for the parameter . Then use this value to find the coordinates of the intersection point.
The angle between a line and a plane can be found using the dot product of the line's direction vector and the plane's normal vector. Remember to use the sine formula for the angle between a line and a plane, .
The reflected beam will also pass through the point of intersection found in part (a). To find its direction, choose another point on the original laser beam, find its reflection across the mirror plane, and then use these two points to determine the direction vector of the reflected beam.
Question 19
HardPaper 2 · calculator20 marksA drone is programmed to follow a straight flight path . The path is described by the Cartesian equation .
Find the vector equation of the drone's flight path , expressing your answer in the form , where .
A ground control station is located at the origin .
Determine the minimum distance from the ground control station to the drone's flight path .
A security laser grid is set up, forming a plane with the equation .
Verify that the drone's flight path lies entirely within the security laser grid .
A second drone is launched from a point . This drone's flight path, , is parallel to the security laser grid and is designed to intersect the -axis.
Find the vector equation of the second drone's flight path , expressing your answer in the form , where .
To convert from Cartesian to vector form, set the given expression equal to a parameter . Then, express , , and in terms of to find the position vector and the direction vector .
Consider the vector from the origin to a general point on the line. What condition must this vector satisfy for the distance to be minimal? Alternatively, you can minimize the squared distance function.
For a line to lie within a plane, two conditions must be met: the line must be parallel to the plane, and at least one point on the line must lie on the plane.
Recall that a line parallel to a plane has its direction vector orthogonal to the plane's normal vector. Also, consider the coordinates of a point on the -axis.
Question 20
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.
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