Intersection and angles between lines & planes: notes and practice questions
- Lines can be represented in vector, parametric, or Cartesian form, using a position vector and a direction vector.
- Planes can be represented in vector, scalar product, or Cartesian form.
- The scalar product form of a plane uses a normal vector, which is perpendicular to the plane.
- The Cartesian form of a plane directly provides the components of the normal vector.
How it is examined
The line-and-plane angle is the one that is systematically got wrong, because the scalar product gives the angle to the normal and the answer needs minus that. Three-plane configurations ask for a geometrical description in words as well as the algebra. 6 to 9 marks across parts, Paper 1.
- Intersections of: a line with a plane; two planes; three planes.
- Angle between: a line and a plane; two planes.
Linking questions
- The geometric half of AHL 1.16. Three planes meeting in a point, in a line, in nothing, or in a sheaf, are the same three outcomes as a linear system.
Practice questions
14 questions · 6 medium · 8 hardQuestion 1
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 2
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 3
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 4
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 5
MediumPaper 1 · no calculator7 marksThe path of a particle is modelled by the line with vector equation . The particle collides with a flat surface modelled by the plane with equation . The point of collision is . Find the coordinates of .
The particle starts its path at the point . Find the shortest distance from the starting point to the surface .
The point of collision lies on both the line and the plane . Express the coordinates of a general point on the line in terms of the parameter , and then substitute these expressions into the equation of the plane.
Recall the formula for the shortest distance from a point to a plane . The formula is given in the formula booklet.
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 · 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 8
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 9
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 10
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 11
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 12
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 13
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 14
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.
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