Difficulty: Intermediate | Prerequisites: MATH 124/125 (single-variable calculus), basic vector operations, dot product, cross product
Planes in 3D are defined by normal vectors, and you can find angles between planes, lines of intersection, and closest points using dot products and projections. Particle motion problems work backwards from acceleration to velocity to position by integrating and using initial or boundary conditions to pin down the constants.
Normal vector
A vector perpendicular to a plane. For the plane ax + by + cz = d, the normal vector is n =
Dot product
For vectors u =
Angle between two planes
The acute angle between two planes equals the acute angle between their normal vectors. You find it using the dot product formula: cos(theta) = |n1 . n2| / (|n1| |n2|).
Parametric equations (of a line)
A way to describe a line in 3D by expressing x, y, and z each as a function of a single parameter t. The form is x = x0 + at, y = y0 + bt, z = z0 + ct, where
Projection of a point onto a plane
The closest point on a plane to a given external point. You find it by dropping a perpendicular from the point to the plane along the normal vector direction. Think of it as the shadow of the point cast straight down onto the plane.
Speed
The magnitude of the velocity vector: speed = |v(t)|. Speed is a scalar (just a number), while velocity is a vector (has direction).
Every plane equation of the form ax + by + cz = d hands you its normal vector for free: n =
For the plane 8x - 4y + z = 3, the normal vector is n = <8, -4, 1>.
For the plane x = 4 (which is really 1x + 0y + 0z = 4), the normal vector is n = <1, 0, 0>.
To find the acute angle between two planes, use the angle between their normals:
cos(theta) = |n1 . n2| / (|n1| |n2|)
The absolute value in the numerator ensures you get the acute angle. Without it, you might get the obtuse supplement.
Worked example (from the exam): Planes P: 8x - 4y + z = 3 and x = 4.
n1 = <8, -4, 1>, n2 = <1, 0, 0>
n1 . n2 = 8
|n1| = sqrt(64 + 16 + 1) = sqrt(81) = 9
|n2| = 1
cos(theta) = 8/9
theta = arccos(8/9)
When two planes intersect, they form a line. To find parametric equations for that line, you need two things: a point on the line, and a direction vector.
Finding the direction vector: The direction of the intersection line is perpendicular to both normal vectors. Use the cross product: d = n1 x n2.
Alternatively, if one of the planes is a coordinate plane (like the xz-plane, where y = 0), you can substitute directly into the other plane's equation and let one variable be the free parameter.
Worked example (from the exam): Intersection of P: 8x - 4y + z = 3 and the xz-plane (y = 0).
Set y = 0 in the plane equation: 8x + z = 3, so z = 3 - 8x.
Let x = t (the free parameter).
Then: x = t, y = 0, z = 3 - 8t.
These are valid parametric equations. You could also set x = 0 to get a specific point (0, 0, 3) and read off the direction vector <1, 0, -8>.
To find the point on a plane P closest to an external point Q, you project Q onto the plane along the normal vector direction.
Method:
Write the line through Q in the direction of n: r(t) = Q + t * n.
Find the value of t where this line meets the plane (substitute into the plane equation).
Plug that t back in to get the closest point.
Worked example (from the exam): Closest point on P: 8x - 4y + z = 3 to the point Q = (26, -7, 10).
The normal is n = <8, -4, 1>.
Parametric line: (26 + 8t, -7 - 4t, 10 + t).
Substitute into 8x - 4y + z = 3:
8(26 + 8t) - 4(-7 - 4t) + (10 + t) = 3
208 + 64t + 28 + 16t + 10 + t = 3
246 + 81t = 3
81t = -243
t = -3
Closest point: (26 + 8(-3), -7 - 4(-3), 10 + (-3)) = (2, 5, 7).
When given acceleration a(t), you recover velocity by integrating, and position by integrating again. Each integration introduces a constant vector C that must be determined from initial or boundary conditions.
v(t) = integral of a(t) dt + C1
r(t) = integral of v(t) dt + C2
The condition "the particle is in the same position at t = 1 and t = 4" means r(1) = r(4). This gives you equations to solve for the components of C1.
Speed at a given time is the magnitude |v(t)|.
Worked example (from the exam): a(t) = <2t, 4, 45 sqrt(t)>, with r(1) = r(4). Find speed at t = 0.
Integrate a(t) to get v(t):
v(t) =
Integrate v(t) to get r(t):
r(t) =
Apply r(1) = r(4), component by component:
x: 1/3 + c1 + d1 = 64/3 + 4c1 + d1, so -3c1 = 63/3 = 21, giving c1 = -7
y: 2 + c2 + d2 = 32 + 4c2 + d2, so -3c2 = 30, giving c2 = -10
z: 12 + c3 + d3 = 12(32) + 4c3 + d3 = 384 + 4c3 + d3, so -3c3 = 372, giving c3 = -124
v(0) = <0 + (-7), 0 + (-10), 0 + (-124)> = <-7, -10, -124>
Speed = |v(0)| = sqrt(49 + 100 + 15376) = sqrt(15525) = sqrt(25 * 621) = 5*sqrt(621)
Angle between planes: cos(theta) = |n1 . n2| / (|n1| |n2|)
Line of intersection (substitution method): Set the constraining variable to 0 (or its fixed value), solve for the remaining variables in terms of a parameter t.
Line of intersection (cross product method): Direction vector d = n1 x n2, then find any point satisfying both plane equations.
Closest point on plane to external point Q: Parametrise the line r(t) = Q + t*n, substitute into the plane equation, solve for t.
Distance from a point to a plane: Dist = |ax0 + by0 + cz0 - d| / sqrt(a^2 + b^2 + c^2)
Integrating acceleration: v(t) = integral a(t) dt + C1, then r(t) = integral v(t) dt + C2.
Integral of t^n: integral t^n dt = t^(n+1)/(n+1) + C (works for all n except -1).
Speed: |v(t)| = sqrt(vx^2 + vy^2 + vz^2)
Students often forget to take the absolute value of the dot product when finding the acute angle between planes. Without it, you may end up with the obtuse angle instead.
When finding the line of intersection with a coordinate plane, students sometimes try the cross-product method when simple substitution (setting y = 0, for example) is much faster and less error-prone.
A common mistake with projection problems is projecting onto the normal vector itself rather than finding the actual point on the plane. The projection formula for distance is not the same as finding the closest point.
In particle-motion problems, students frequently forget that each component of the velocity vector gets its own constant of integration, giving three separate constants, not one shared constant.
Plane-normal-vector problems appear on nearly every MATH 126 final. Know how to read the normal from the equation instantly.
The angle-between-planes question is a reliable 4-point freebie if you remember the absolute value.
Closest-point-on-a-plane is a projection problem that also tests your comfort with parametric lines. It frequently appears as part (c) of a multi-part planes question.
Particle-motion problems that give you a boundary condition like r(a) = r(b) are testing whether you can set up the algebra correctly after integrating. The integration itself is straightforward; the exam difficulty is in using the condition to solve for constants.
True or False: The normal vector to the plane 3x + 2y - z = 7 is <3, 2, 1>.
Fill in the blank: To find the acute angle between two planes, take the ______ of the dot product of their normals before dividing.
True or False: Speed and velocity are the same thing.
Fill in the blank: Integrating the acceleration vector gives the ______ vector, plus a constant.
True or False: The closest point on a plane to an external point lies along the line through the external point in the direction of the plane's normal vector.
Answers: 1. False (it should be <3, 2, -1>). 2. Absolute value. 3. False (speed is a scalar, velocity is a vector). 4. Velocity. 5. True.
Q: The plane P has equation 2x - 3y + 6z = 12. What is the acute angle between P and the xy-plane?
A: Normal to P is <2, -3, 6>. Normal to the xy-plane is <0, 0, 1>. Dot product = 6. |n1| = sqrt(4+9+36) = 7, |n2| = 1. cos(theta) = 6/7, so theta = arccos(6/7).
Q: Write parametric equations for the line of intersection of 2x + y - z = 4 and the yz-plane.
A: The yz-plane is x = 0. Set x = 0: y - z = 4. Let z = t, then y = 4 + t. Parametric equations: x = 0, y = 4 + t, z = t.
Q: A particle has acceleration a(t) = <6t, 2, 0> and initial velocity v(0) = <1, -3, 5>. What is the speed at t = 2?
A: v(t) = <3t^2 + c1, 2t + c2, c3>. From v(0) = <1, -3, 5>: c1 = 1, c2 = -3, c3 = 5. So v(2) = <13, 1, 5>. Speed = sqrt(169 + 1 + 25) = sqrt(195).
Q: Find the point on the plane x + 2y + 2z = 9 closest to the origin.
A: Normal n = <1, 2, 2>. Line: r(t) = t<1, 2, 2>. Substitute: t + 4t + 4t = 9, so 9t = 9, t = 1. Closest point: (1, 2, 2).
Normal vectors reappear in MATH 126 when you study tangent planes to surfaces (Problem 9 on this exam). The formula for the tangent plane at a point on z = f(x,y) uses the gradient as a generalised normal vector.
The projection technique used for closest-point problems is the same idea behind Lagrange multipliers and constrained optimisation: you are finding where a perpendicular condition is satisfied.
Particle motion connects directly to arc length, curvature, and the TNB frame later in the course. The relationship between acceleration, velocity, and position via integration is also the foundation for setting up centre-of-mass problems (Problem 5 on this exam).
normal vector, plane equation, angle between planes, dot product, cross product, parametric equations, line of intersection, coordinate planes, xz-plane, yz-plane, closest point on plane, projection onto plane, distance from point to plane, acceleration vector, velocity vector, position vector, integrating acceleration, speed vs velocity, boundary condition r(a) = r(b), MATH 126, Calculus III, University of Washington, multivariable calculus, 3D geometry