Difficulty: Intermediate | Prerequisites: Basic vector operations (dot product, cross product, vector magnitude)
Any vector can be split into a component along a plane's normal and a component lying in the plane. Flipping the normal component gives you the reflection; flipping the in-plane component gives the ricochet. These decompositions also let you compute angles between a vector and its reflected image using the dot product formula.
Scalar projection (comp)
The signed length of the shadow one vector casts along another: (P · n) / ||n||. In simple terms, it tells you "how much" of P points in the n direction.
Vector projection (proj)
The vector component of P that lies along n: proj_n P = ((P · n) / ||n||)(n / ||n||). Think of it as the shadow of P cast onto the line defined by n, returned as a vector, not just a number.
Orthogonal projection (orth)
The leftover after you subtract the projection: orth_n P = P – proj_n P. This is the part of P that lies in the plane perpendicular to n.
Reflection vector (refl)
The mirror image of P through the plane: refl_n P = –proj_n P + orth_n P. You negate the normal component and keep the in-plane component.
Ricochet vector (ric)
The direction a billiard ball takes after hitting the plane: ric_n P = proj_n P – orth_n P = –refl_n P. You keep the normal component and negate the in-plane component.
Normal vector
A vector perpendicular to a surface or plane. For the xz-plane the normal is (0, 1, 0); for the plane z = x + y, rewrite as x + y – z = 0, so the normal is (1, 1, −1).
Orthogonal vectors
Two vectors whose dot product is zero. They meet at a right angle.
Every vector P can be written as the sum of two perpendicular pieces:
Normal component (along n): proj_n P = ((P · n) / (n · n)) n
In-plane component: orth_n P = P – proj_n P
This decomposition is the foundation for reflection, ricochet, and angle calculations.
Let P = (1, 2, 1). The xz-plane has normal n = (0, 1, 0).
proj_n P = ((1, 2, 1) · (0, 1, 0) / 1) × (0, 1, 0) = (0, 2, 0)
orth_n P = (1, 2, 1) – (0, 2, 0) = (1, 0, 1)
refl_n P = –(0, 2, 0) + (1, 0, 1) = (1, −2, 1)
Geometric check: reflecting through the xz-plane should negate only the y-component, which is exactly what happened.
Same point P = (1, 2, 1). Rewrite the plane as x + y – z = 0, so n = (1, 1, −1).
P · n = 1 + 2 – 1 = 2, and ||n||² = 3
proj_n P = (2/3)(1, 1, −1) = (2/3, 2/3, −2/3)
orth_n P = (1, 2, 1) – (2/3, 2/3, −2/3) = (1/3, 4/3, 5/3)
refl_n P = –(2/3, 2/3, −2/3) + (1/3, 4/3, 5/3) = (−1/3, 2/3, 7/3)
Both P and refl_n P have magnitude √6. Their dot product is 10/3.
cos θ = (P · refl_n P) / (||P|| × ||refl_n P||) = (10/3) / 6 = 5/9
θ = cos⁻¹(5/9)
Note that the angle between P and its reflection equals twice the angle between P and the plane.
A billiard ball travelling along –P hits the plane at the origin. The ricochet direction must:
Lie in the plane defined by P and n
Stay on the same side of the plane as P
Make the same angle with the plane as P
The unique vector satisfying all three is ric_n P = –refl_n P = proj_n P – orth_n P.
These come directly from the kind of true/false questions that appear on MATH 215 midterms. Work through each before reading the answer.
If v and u are unit vectors, is v + u orthogonal to v – u?
TRUE. Compute (v + u) · (v – u) = v · v – u · u = ||v||² – ||u||² = 1 – 1 = 0. The equal-magnitude condition is what makes this work.
If u and v are unit vectors in R³, is c(t) = cos(t) u + sin(t) v always a circle?
FALSE. It traces a circle only when u and v are orthogonal. If u and v are not perpendicular, the curve is an ellipse (and if u = v, it collapses to a line segment).
Is d/dt |c(t)| = |c'(t)| for a differentiable vector function?
FALSE. The derivative of the magnitude is (c(t) · c'(t)) / |c(t)|, which equals the component of c' in the direction of c, not the full magnitude of c'. They are equal only when c(t) and c'(t) are parallel.
Does c(t) = (2 + t³, 2 – t³, 8t³) trace a straight line on a hyperbolic paraboloid?
TRUE. The direction vector is (1, −1, 8) (factor out t³), so it is a line through (2, 2, 0). Substituting x = 2 + t³ and y = 2 – t³ into z = xy – 4 (a hyperbolic paraboloid) gives z = 8t³, confirming the curve lies on it.
If g(x) = f(x, 0) and h(y) = f(0, y) are both differentiable at 0, is f differentiable at (0, 0)?
FALSE. Differentiability of slices along the axes does not guarantee differentiability of the full function. A classic counterexample is f(x, y) = xy / (x² + y²) (with f(0,0) = 0), whose x- and y-slices through the origin are both zero (hence differentiable), but f is not even continuous at the origin along other paths.
Do A = (1, 2, 2) and B = (2, 2, 1) lie on the same side of x + y – z = 2?
FALSE. Plug each point into x + y – z – 2: for A you get 1 + 2 – 2 – 2 = −1 (negative), for B you get 2 + 2 – 1 – 2 = 1 (positive). Different signs mean different sides.
Does constant curvature imply a constant tangent vector?
FALSE. A circle has constant curvature but its tangent vector rotates continuously. Constant curvature means the rate of turning is steady, not that the direction is fixed.
\text{proj}_{\vec{n}} \vec{P} = \frac{\vec{P} \cdot \vec{n}}{\vec{n} \cdot \vec{n}} \, \vec{n}\text{orth}_{\vec{n}} \vec{P} = \vec{P} - \text{proj}_{\vec{n}} \vec{P}\text{refl}_{\vec{n}} \vec{P} = -\text{proj}_{\vec{n}} \vec{P} + \text{orth}_{\vec{n}} \vec{P} = \vec{P} - 2\,\text{proj}_{\vec{n}} \vec{P}\text{ric}_{\vec{n}} \vec{P} = \text{proj}_{\vec{n}} \vec{P} - \text{orth}_{\vec{n}} \vec{P} = -\text{refl}_{\vec{n}} \vec{P}\cos\theta = \frac{\vec{P} \cdot \text{refl}_{\vec{n}} \vec{P}}{\|\vec{P}\| \, \|\text{refl}_{\vec{n}} \vec{P}\|}Note: the compact form refl_n P = P – 2 proj_n P is often the fastest way to compute a reflection in one step.
Students often divide by ||n|| twice in the projection formula when they should divide by ||n||² (equivalently, by n · n). The scalar projection divides once; the vector projection divides twice, but one of those divisions is baked into the unit vector n/||n||.
Reflection and ricochet are frequently swapped. Remember: reflection negates the normal component; ricochet negates the in-plane component. They are negatives of each other.
When finding the normal to a plane like z = x + y, students sometimes use n = (0, 0, 1) (the normal to z = const). Rewrite the equation as x + y – z = 0 first, and read the coefficients: n = (1, 1, −1).
The statement d/dt |c(t)| = |c'(t)| looks plausible but is false in general. A common trap on true/false sections.
⚠️ Projection, orthogonal decomposition, and reflection are tested almost every term on MATH 215 midterms, both as computation and as conceptual true/false.
⚠️ The multi-step structure (project, then orthogonalise, then reflect) is a template: exams frequently hand you a new plane or a twist (like the ricochet) and ask you to adapt the same three steps.
⚠️ Angle-between-vectors problems using the dot product formula appear regularly. Be comfortable going from the dot product to cos⁻¹.
⚠️ True/false sections reward careful counterexample thinking. "Differentiable slices imply differentiable function" and "constant curvature implies constant tangent" are recurring traps.
True or false: proj_n P and orth_n P are always perpendicular to each other.
Fill in the blank: refl_n P = P – 2 × ______.
True or false: if ||P|| = ||refl_n P||, something has gone wrong in your calculation.
The normal vector to the plane 3x – y + 2z = 7 is ______.
True or false: (v + u) · (v – u) = 0 for any two vectors of equal magnitude.
Answers: 1. True. 2. proj_n P. 3. False (they always have the same magnitude). 4. (3, −1, 2). 5. True.
Q: Reflect the point (3, 1, 2) through the plane 2x – y + z = 0.
A: n = (2, −1, 1). P · n = 6 – 1 + 2 = 7. ||n||² = 6. proj_n P = (7/6)(2, −1, 1) = (7/3, −7/6, 7/6). refl_n P = P – 2 proj_n P = (3 – 14/3, 1 + 7/3, 2 – 7/3) = (−5/3, 10/3, −1/3).
Q: Two unit vectors satisfy u · v = 1/2. What is |u + v|?
A: |u + v|² = u · u + 2(u · v) + v · v = 1 + 1 + 1 = 3. So |u + v| = √3.
Q: A billiard ball travels along –(1, 0, 1) and hits the plane y = 0 at the origin. In what direction does it ricochet?
A: P = (1, 0, 1), n = (0, 1, 0). proj_n P = (0, 0, 0), orth_n P = (1, 0, 1). ric = proj – orth = –(1, 0, 1). The ball bounces straight back because it was travelling entirely within the plane.
Q: Find the angle between (1, 1, 0) and its reflection through the plane x – y = 0.
A: n = (1, −1, 0). proj_n P = (1)(1, −1, 0)/2 = (1/2, −1/2, 0). refl = P – 2 proj = (0, 2, 0). cos θ = ((1,1,0) · (0,2,0)) / (√2 × 2) = 2/(2√2) = 1/√2. θ = π/4.
Projection onto a normal vector reappears when you compute tangent planes to surfaces (the gradient of a level surface gives the normal). The same decomposition is used in directional derivatives, where you project the gradient onto a unit direction.
The dot product and angle formula underpin the definition of curvature and the analysis of space curves later in the course.
Related Terms / Search Tags
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