This deck covers lines and planes in space and the distances between them, then cylinders, the six quadric surfaces, and the cylindrical and spherical coordinate systems. It targets the confusion between a normal vector and a direction vector, the sign pattern that separates a one-sheet surface from a two-sheet one, and the absolute-value error in the distance-to-a-plane formula.
Subject: Calculus III · 109 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Write the parametric and symmetric equations of a line in space from a point and a direction vector.
2. Write the equation of a plane from a point and a normal vector, and from three points.
3. Find the angle between two planes, and the distance from a point to a plane and to a line.
4. Identify and trace cylinders and the six quadric surfaces.
5. Convert between rectangular, cylindrical, and spherical coordinates.
Warm-up
Discussion prompt
Before we open Week 2 - Geometry of Space: without looking back, what was the main idea of Week 1 - Vectors & Their Properties, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck distinguishes vectors from scalars and covers component form and position vectors in two and three dimensions, then sets up the space coordinate system with distance and spheres. It works through addition, scaling, magnitude, unit vectors, and the standard basis, then the dot product and its uses for angle, orthogonality, projections, and work, and the cross product in determinant form, with the right-hand rule, area, and the triple scalar product for volume. It targets the classic traps: mixing up the scalar dot product with the vector cross product, forgetting that the cross product is not commutative, skipping normalization, and reporting a scalar projection when a vector was asked for.
Concept
A single point does not pin down a line -- infinitely many lines pass through it. You also need to say which way the line goes.
So a line in space is fixed by two things: one point it passes through, and a direction vector that points along it.
direction vector — A nonzero vector parallel to the line. Any scalar multiple of it works equally well, since a longer or shorter arrow points the same way along the line.
Counterexample
Discussion prompt
A single point does not pin down a line -- infinitely many lines pass through it. You also need to say which way the line goes.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Intuition
Think of a walker who starts at a fixed point and marches in a straight, fixed direction forever, forwards and backwards.
The parameter is like time. At each instant the walker is at a new spot on the line. Negative time is walking backwards.
Every point of the line is start plus some amount of the direction.
Analogy
Discussion prompt
Explain Start here, walk that way by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of a walker who starts at a fixed point and marches in a straight, fixed direction forever, forwards and backwards.
Concept
Let the line pass through a point with position vector shown below, with direction vector as shown.
\[ \mathbf{r}_0 = \langle x_0, y_0, z_0\rangle, \qquad \mathbf{v} = \langle a, b, c\rangle \]
Every point on the line is the start plus a scalar multiple of the direction:
\[ \mathbf{r}(t) = \mathbf{r}_0 + t\,\mathbf{v} \]
As the parameter runs over all real numbers, the tip of this vector sweeps out the whole line.
Explain it
Discussion prompt
Explain Vector equation of a line to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Every point on the line is the start plus a scalar multiple of the direction:
Concept
Reading the vector equation one coordinate at a time gives three scalar equations, the parametric equations:
\[ x = x_0 + a t, \quad y = y_0 + b t, \quad z = z_0 + c t \]
The point supplies the constants; the direction vector supplies the coefficients of the parameter.
Concept
If none of the direction components is zero, solve each parametric equation for the parameter and set the results equal. This eliminates the parameter:
\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \]
The denominators are exactly the direction components. If one component is zero, that variable is simply held constant and written as a separate equation.
Ranking
Put in order
Put the moves of Line through two points into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The vector from P to Q lies along the line, so it is a valid direction vector.
Worked example
Find parametric and symmetric equations of the line through these two points.
\[ P = (2, -1, 3), \qquad Q = (5, 1, 4) \]
Build a direction vector from the two points
Why: The vector from P to Q lies along the line, so it is a valid direction vector.
\[ \mathbf{v} = Q - P = \langle 3, 2, 1\rangle \]
Use P as the base point in the parametric form
Why: Any point on the line works; P is given, so use it with the direction components as coefficients.
\[ x = 2 + 3t, \quad y = -1 + 2t, \quad z = 3 + t \]
Solve each for the parameter to get symmetric form
Why: All three direction components are nonzero, so the parameter can be eliminated cleanly.
\[ \frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{1} \]
Verify by plugging in the endpoints
Why: At the parameter value 0 the point is (2, -1, 3) = P, and at 1 it is (5, 1, 4) = Q. Both given points lie on the line, so the equations are correct.
\[ t=0:\;(2,-1,3)=P \quad t=1:\;(5,1,4)=Q \]
Picture it
Animation
Shows: Each line of the worked example "Line through two points", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the parameter value 0 the point is (2, -1, 3) = P, and at 1 it is (5, 1, 4) = Q. Both given points lie on the line, so the equations are correct.
Concept
Two lines in space can be parallel (directions are scalar multiples), intersecting (they share a point), or skew (neither parallel nor meeting).
Skew lines are the genuinely three-dimensional case -- they never touch yet are not parallel, like an overpass crossing above a road.
First compare directions. If they are not parallel, then test whether the lines actually share a common point.
Step zero
Discussion prompt
Are these two lines parallel? — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Read off each direction vector
Answer:
Worked example
Decide whether the two lines below are parallel.
\[ L_1:\ \langle 1+2t,\ 3-t,\ 2+4t\rangle \]
\[ L_2:\ \langle 4-4s,\ 1+2s,\ 6-8s\rangle \]
Read off each direction vector
Why: The coefficients of the parameter in each coordinate form the direction vector.
\[ \mathbf{v}_1 = \langle 2, -1, 4\rangle, \quad \mathbf{v}_2 = \langle -4, 2, -8\rangle \]
Test whether one direction is a scalar multiple of the other
Why: Parallel lines have parallel directions, meaning one vector equals a constant times the other.
\[ \mathbf{v}_2 = -2\,\mathbf{v}_1 \]
Verify component by component
Why: Multiplying the first direction by -2 gives exactly the second direction in every coordinate, so the lines are parallel.
\[ -2\langle 2,-1,4\rangle = \langle -4, 2, -8\rangle \]
Picture it
Animation
Shows: Each line of the worked example "Are these two lines parallel?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Multiplying the first direction by -2 gives exactly the second direction in every coordinate, so the lines are parallel.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A plane is given as the equation below, and someone is asked for a line perpendicular to it. They grab a vector lying in the plane and call it the line's direction.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: A vector inside the plane runs along the surface, not perpendicular to it.
The coefficients of a plane equation are its normal vector, which points perpendicular to the plane. That is exactly the direction a perpendicular line needs.
Why: A vector inside the plane runs along the surface, not perpendicular to it. The resulting line is parallel to the plane, not perpendicular.
Trap
A plane is given as the equation below, and someone is asked for a line perpendicular to it. They grab a vector lying in the plane and call it the line's direction.
\[ 2x - y + 4z = 12 \]
Wrong: uses an in-plane vector as the direction
Why: A vector inside the plane runs along the surface, not perpendicular to it. The resulting line is parallel to the plane, not perpendicular.
\[ \mathbf{v}_{\text{wrong}} = \langle 1, 2, 0\rangle \ (\text{lies in the plane}) \]
The coefficients of a plane equation are its normal vector, which points perpendicular to the plane. That is exactly the direction a perpendicular line needs.
\[ \mathbf{n} = \langle 2, -1, 4\rangle \]
Right: use the normal as the line's direction
Why: A plane is defined by its normal (perpendicular); a line is defined by its direction (along it). For a line perpendicular to the plane, the plane's normal IS the line's direction.
\[ \mathbf{v}_{\text{right}} = \mathbf{n} = \langle 2, -1, 4\rangle \]
Notation
Annotate
From Trap: normal vector vs direction vector — read this one piece at a time. What is each part doing?
On: \( \mathbf{v}_{\text{wrong}} = \langle 1, 2, 0\rangle \ (\text{lies in the plane}) \)
Pattern
1. Get a point and a direction vector
Why: From two points, subtract to get the direction. From a parallel line, reuse its direction. For a line perpendicular to a plane, use the plane's normal.
2. Write the parametric equations
Why: Each coordinate is a base coordinate plus the parameter times the matching direction component.
3. Eliminate the parameter for symmetric form
Why: Set the three parameter expressions equal, using the direction components as denominators (only where they are nonzero).
Elimination
Eliminate the wrong options
Which set of parametric equations describes this line?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Each coordinate is the point's coordinate plus the parameter times the matching direction component. The y-component of the direction is 0, so y stays fixed at 4. This gives x = 1 + 3t, y = 4, z = -2 + 5t.
Check
Consider the line through the point below with the given direction vector.
\[ P = (1, 4, -2), \qquad \mathbf{v} = \langle 3, 0, 5\rangle \]
Check your understanding
Which set of parametric equations describes this line?
Answer: A
Why: Each coordinate is the point's coordinate plus the parameter times the matching direction component. The y-component of the direction is 0, so y stays fixed at 4. This gives x = 1 + 3t, y = 4, z = -2 + 5t.
Concept
A plane is fixed by one point on it and a normal vector -- a vector perpendicular to the whole plane.
normal vector — A nonzero vector perpendicular to a plane. It points straight out of the surface and controls the plane's tilt; every vector lying in the plane is perpendicular to it.
This is the key contrast with a line: a line is defined by a vector that runs along it, a plane by a vector that stands across it.
Definition probe
Sort into buckets
Every line below is part of the definition of direction vector or of normal vector — one or the other, never both. Put each where it belongs.
Intuition
Picture a flat sheet of glass. The normal vector is a pencil balanced straight up off its surface -- it tells you which way the glass faces.
Tilt the pencil and the glass tilts with it. Fix the pencil's direction and pick one point, and the entire plane is locked in place.
A point on the plane can be reached only by moving in directions perpendicular to the normal. That single fact becomes the plane's equation.
Concept
Take a fixed point and the normal below. A point lies on the plane exactly when the vector from the fixed point to it is perpendicular to the normal.
\[ P_0 = (x_0, y_0, z_0), \qquad \mathbf{n} = \langle a, b, c\rangle \]
Perpendicular means the dot product is zero, which gives the point-normal form:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
Expanding and collecting the constant gives the standard form, whose coefficients are exactly the normal components:
\[ ax + by + cz = d, \qquad d = a x_0 + b y_0 + c z_0 \]
Estimation
Predict first
Find the equation of the plane through the point below with the given normal vector.
Commit before you compute: what does Plane from a point and a normal come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by substituting the point
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Plugging P0 into the left side gives 2(1) - (2) + 4(3) = 2 - 2 + 12 = 12, which matches the right side, so the point lies on the plane.
Worked example
Find the equation of the plane through the point below with the given normal vector.
\[ P_0 = (1, 2, 3), \qquad \mathbf{n} = \langle 2, -1, 4\rangle \]
Write the point-normal form
Why: The normal components are the coefficients, and each variable is shifted by the matching coordinate of the point.
\[ 2(x - 1) - 1(y - 2) + 4(z - 3) = 0 \]
Expand and collect constants
Why: Distribute each coefficient, then combine the number terms on one side.
\[ 2x - y + 4z - 12 = 0 \]
Write the standard form
Why: Move the constant to the right to display the normal coefficients cleanly.
\[ 2x - y + 4z = 12 \]
Verify by substituting the point
Why: Plugging P0 into the left side gives 2(1) - (2) + 4(3) = 2 - 2 + 12 = 12, which matches the right side, so the point lies on the plane.
\[ 2(1) - (2) + 4(3) = 12 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Plane from a point and a normal", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Plugging P0 into the left side gives 2(1) - (2) + 4(3) = 2 - 2 + 12 = 12, which matches the right side, so the point lies on the plane.
Concept
Three points that are not all on one line fix a plane. But the equation needs a normal, and points alone do not give one directly.
Make two vectors from the points -- both lie in the plane. Their cross product is perpendicular to both, so it serves as the normal.
\[ \mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} \]
Then use that normal with any one of the three points in the point-normal form.
Worked example
Find the plane through these three points.
\[ A = (1,0,0), \quad B = (0,2,0), \quad C = (0,0,3) \]
Form two edge vectors from A
Why: Subtracting A from B and from C gives two vectors that both lie in the plane.
\[ \overrightarrow{AB} = \langle -1, 2, 0\rangle, \quad \overrightarrow{AC} = \langle -1, 0, 3\rangle \]
Cross the edge vectors to get a normal
Why: The cross product is perpendicular to both edges, hence normal to the plane they span.
\[ \mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \langle 6, 3, 2\rangle \]
Apply the point-normal form using A
Why: With the normal known, any of the three points completes the equation; A is simplest.
\[ 6(x - 1) + 3(y - 0) + 2(z - 0) = 0 \ \Rightarrow\ 6x + 3y + 2z = 6 \]
Verify all three points satisfy it
Why: A gives 6(1)=6, B gives 3(2)=6, and C gives 2(3)=6. All three points satisfy the equation, so the plane is correct.
\[ A:6,\quad B:6,\quad C:6 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Plane through three points", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A gives 6(1)=6, B gives 3(2)=6, and C gives 2(3)=6. All three points satisfy the equation, so the plane is correct.
Pattern
1. Secure a normal vector
Why: Given directly, take it. Given three points, cross two edge vectors. Given a parallel plane, reuse its normal.
2. Pick any point on the plane
Why: Any known point works with the point-normal form.
3. Write point-normal form and simplify
Why: Multiply the normal components against the shifted variables, expand, and collect the constant into standard form.
Check
Find the plane through the three points below.
\[ A = (0,0,0), \quad B = (1,1,0), \quad C = (0,1,1) \]
Check your understanding
Which equation describes this plane?
Answer: A
Why: The edge vectors are AB = <1,1,0> and AC = <0,1,1>. Their cross product is <1,-1,1>, the normal. Since A is the origin, the constant d is 0, giving x - y + z = 0.
Concept
Two planes meet at an angle. That dihedral angle is the same as the angle between their normal vectors.
Use the dot-product angle formula on the normals. The absolute value keeps the reported angle acute.
\[ \cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{\|\mathbf{n}_1\|\,\|\mathbf{n}_2\|} \]
If the normals are parallel the planes are parallel; if the normals are perpendicular the planes are perpendicular.
Intuition
Open a book slightly. The two covers are planes; the spine is their line of intersection.
Stand a pencil straight up from each cover -- those are the normals. As you open the book wider, the pencils spread by the same angle the covers do.
So measuring the angle between normals is the same as measuring the angle between the planes themselves.
Step zero
Discussion prompt
Angle between two planes — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Read off the normals from the coefficients
Answer:
Worked example
Find the acute angle between these two planes.
\[ x + y + z = 1 \qquad \text{and} \qquad x - y + z = 5 \]
Read off the normals from the coefficients
Why: The coefficients of each plane equation are the components of its normal vector.
\[ \mathbf{n}_1 = \langle 1,1,1\rangle, \quad \mathbf{n}_2 = \langle 1,-1,1\rangle \]
Compute the dot product and magnitudes
Why: These feed the cosine formula for the angle between the normals.
\[ \mathbf{n}_1 \cdot \mathbf{n}_2 = 1 - 1 + 1 = 1, \quad \|\mathbf{n}_1\| = \|\mathbf{n}_2\| = \sqrt{3} \]
Apply the cosine formula
Why: Divide the absolute dot product by the product of the magnitudes.
\[ \cos\theta = \frac{|1|}{\sqrt{3}\cdot\sqrt{3}} = \frac{1}{3} \]
Verify the result is a valid acute angle
Why: Since one-third is between 0 and 1, the inverse cosine gives an acute angle of about 70.5 degrees, which is a sensible angle between two planes.
\[ \theta = \arccos\tfrac{1}{3} \approx 70.5^{\circ} \]
Picture it
Animation
Shows: Each line of the worked example "Angle between two planes", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Since one-third is between 0 and 1, the inverse cosine gives an acute angle of about 70.5 degrees, which is a sensible angle between two planes.
Concept
The distance from a point to a plane is measured straight along the normal -- the shortest possible path.
For the plane written in standard form and a point, the distance is:
\[ D = \frac{|a x_1 + b y_1 + c z_1 - d|}{\sqrt{a^2 + b^2 + c^2}} \]
The numerator measures how far the point is from satisfying the equation; the denominator is the length of the normal, which rescales that gap into true distance.
Intuition
Drop a vector from any point on the plane to the outside point. Its shadow along the normal direction is the distance.
Projecting onto the normal is a dot product; dividing by the normal's length turns that into a length. That is exactly the formula.
Because distance can never be negative, the numerator is wrapped in an absolute value.
Hypothesis
Predict first
Distance from a point to a plane is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Identify the normal and constant
Why: The coefficients give the normal, and the right-hand side is the constant d.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Find the distance from the point to the plane below.
\[ P = (1, 2, 3), \qquad 2x - 2y + z = 6 \]
Identify the normal and constant
Why: The coefficients give the normal, and the right-hand side is the constant d.
\[ \mathbf{n} = \langle 2, -2, 1\rangle, \quad d = 6 \]
Evaluate the numerator with absolute value
Why: Substitute the point into ax + by + cz - d, then take the absolute value so the distance is nonnegative.
\[ |2(1) - 2(2) + 1(3) - 6| = |{-5}| = 5 \]
Divide by the length of the normal
Why: The magnitude of the normal rescales the numerator into true distance.
\[ \|\mathbf{n}\| = \sqrt{4 + 4 + 1} = 3, \qquad D = \frac{5}{3} \]
Verify the answer is positive and reasonable
Why: The distance five-thirds is positive, as every genuine distance must be, confirming the absolute value was applied correctly.
\[ D = \tfrac{5}{3} > 0 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Distance from a point to a plane", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The distance five-thirds is positive, as every genuine distance must be, confirming the absolute value was applied correctly.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Using the same point and plane, a student plugs in but forgets the absolute value bars in the numerator.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Without the absolute value the numerator is -5, giving a negative result.
Wrap the numerator in absolute value bars before dividing. The sign only tells you which side of the plane the point is on.
Why: Without the absolute value the numerator is -5, giving a negative result. A distance can never be negative, so this answer is impossible.
Trap
Using the same point and plane, a student plugs in but forgets the absolute value bars in the numerator.
\[ P = (1,2,3), \qquad 2x - 2y + z = 6 \]
Wrong: reports a negative distance
Why: Without the absolute value the numerator is -5, giving a negative result. A distance can never be negative, so this answer is impossible.
\[ D_{\text{wrong}} = \frac{2(1) - 2(2) + 3 - 6}{3} = \frac{-5}{3} = -\tfrac{5}{3} \]
Wrap the numerator in absolute value bars before dividing. The sign only tells you which side of the plane the point is on.
Right: take the absolute value first
Why: The absolute value of -5 is 5, so the distance is five-thirds, a positive number as required.
\[ D_{\text{right}} = \frac{|{-5}|}{3} = \frac{5}{3} \]
Translation
\( D_{\text{wrong}} = \frac{2(1) - 2(2) + 3 - 6}{3} = \frac{-5}{3} = -\tfrac{5}{3} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Pattern
1. Put the plane in standard form
Why: You need the coefficients a, b, c and the constant d clearly identified.
2. Plug the point into ax + by + cz - d and take the absolute value
Why: This measures how far the point misses the equation; the absolute value keeps distance nonnegative.
3. Divide by the magnitude of the normal
Why: The square root of the sum of squares of the coefficients converts the gap into true perpendicular distance.
Elimination
Eliminate the wrong options
What is the distance from the point to the plane?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The numerator is |2(3) + 1 - 2(2) - 4| = |6 + 1 - 4 - 4| = |-1| = 1. The normal length is the square root of (4 + 1 + 4) = 3. So the distance is 1/3.
Check
Find the distance from the point to the plane.
\[ P = (3, 1, 2), \qquad 2x + y - 2z = 4 \]
Check your understanding
What is the distance from the point to the plane?
Answer: A
Why: The numerator is |2(3) + 1 - 2(2) - 4| = |6 + 1 - 4 - 4| = |-1| = 1. The normal length is the square root of (4 + 1 + 4) = 3. So the distance is 1/3.
Concept
For a line through a point Q with direction vector, and an external point P, the distance uses a cross product.
\[ D = \frac{\|\overrightarrow{QP} \times \mathbf{v}\|}{\|\mathbf{v}\|} \]
The cross product's magnitude is the area of the parallelogram spanned by the two vectors; dividing by the base length leaves the height, which is the distance.
Intuition
The vector from the line to the point, together with the direction vector, spans a parallelogram.
A parallelogram's area equals base times height. The base is the direction's length; the height is the perpendicular distance we want.
So divide the area (the cross-product magnitude) by the base (the direction's length) to recover the height.
Ranking
Put in order
Put the moves of Distance from a point to a line into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Since Q is the origin, this vector is just the coordinates of P.
Worked example
Find the distance from the point P to the line through the origin with the given direction.
\[ P = (3, 1, -1), \quad Q = (0,0,0), \quad \mathbf{v} = \langle 1, 2, 2\rangle \]
Form the vector from the line to the point
Why: Since Q is the origin, this vector is just the coordinates of P.
\[ \overrightarrow{QP} = \langle 3, 1, -1\rangle \]
Compute the cross product with the direction
Why: The cross product is perpendicular to both vectors, and its length gives the parallelogram area.
\[ \overrightarrow{QP} \times \mathbf{v} = \langle 4, -7, 5\rangle \]
Divide the two magnitudes
Why: The cross-product length over the direction length gives the perpendicular height.
\[ D = \frac{\sqrt{16+49+25}}{\sqrt{1+4+4}} = \frac{\sqrt{90}}{3} = \sqrt{10} \]
Verify the cross product is perpendicular to the direction
Why: The dot product of <4,-7,5> with <1,2,2> is 4 - 14 + 10 = 0, confirming the cross product was computed correctly.
\[ \langle 4,-7,5\rangle \cdot \langle 1,2,2\rangle = 4 - 14 + 10 = 0 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Distance from a point to a line", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The dot product of <4,-7,5> with <1,2,2> is 4 - 14 + 10 = 0, confirming the cross product was computed correctly.
Concept
A cylinder is the surface swept out by moving a straight line (a ruling) along a fixed plane curve, keeping the line parallel to a fixed axis.
The giveaway in an equation: one variable is missing. That missing variable is the axis the surface runs along, free to be anything.
The equation below has no z, so it is a circular cylinder of radius 2 running along the z-axis.
\[ x^2 + y^2 = 4 \]
Intuition
Draw the curve in the plane where all its variables live, then drag that same curve straight up and down the missing axis.
Because the missing variable never appears, the cross-section looks identical at every height. The curve is simply copied along the axis.
Estimation
Predict first
Describe the surface given below in three-dimensional space.
Commit before you compute: what does Identify and describe a cylinder come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the cross-sections are identical
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Setting y to any constant leaves the same parabola z = x squared, so the parabola is copied along the y-axis, confirming a parabolic cylinder.
Worked example
Describe the surface given below in three-dimensional space.
\[ z = x^2 \]
Spot the missing variable
Why: The variable y does not appear, so the surface is a cylinder whose rulings run parallel to the y-axis.
Identify the generating curve
Why: In the xz-plane the equation is a parabola, so this is a parabolic cylinder.
\[ z = x^2 \ \text{(a parabola in the } xz\text{-plane)} \]
Verify the cross-sections are identical
Why: Setting y to any constant leaves the same parabola z = x squared, so the parabola is copied along the y-axis, confirming a parabolic cylinder.
\[ y = k \ \Rightarrow\ z = x^2 \ \text{for every } k \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Identify and describe a cylinder", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Setting y to any constant leaves the same parabola z = x squared, so the parabola is copied along the y-axis, confirming a parabolic cylinder.
Concept
A trace is the curve you get by intersecting a surface with a coordinate plane, found by setting one variable to zero (or to a constant).
trace — The intersection of a surface with a plane. Setting a variable to a constant collapses the surface equation to a two-variable curve you can recognize.
Reading a few traces -- one horizontal and a couple vertical -- tells you the whole shape of a surface.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of direction vector, normal vector, trace as Week 2 - Geometry of Space uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
Slicing a surface with parallel planes is like cutting a loaf of bread: each slice shows a cross-section.
If every horizontal slice is an ellipse that grows then shrinks, you are looking at something rounded and closed. If slices keep growing, the surface opens up forever.
The pattern of the slices is the identity of the surface.
Concept
A quadric surface is the graph of a second-degree equation in three variables. There are six standard types.
\[ Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0 \]
Two features sort them out: which squared terms are present, and the pattern of plus and minus signs when the equation is written in standard form.
Concept
All three variables are squared, all with plus signs, equal to one. It is a closed, bounded surface, like a stretched sphere.
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \]
Every trace -- horizontal or vertical -- is an ellipse. The surface reaches only as far as the intercepts on each axis.
Step zero
Discussion prompt
Trace an ellipsoid — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Classify by signs
Answer:
Worked example
Identify and trace the surface below.
\[ \frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{25} = 1 \]
Classify by signs
Why: Three squared terms, all positive, equal to one, is the signature of an ellipsoid.
Take the trace in the xy-plane
Why: Setting z to zero collapses the equation to a two-variable curve.
\[ z=0:\ \frac{x^2}{4} + \frac{y^2}{9} = 1 \ \text{(ellipse)} \]
Take the other two traces
Why: Setting x to zero and y to zero each gives another ellipse, confirming a closed rounded surface.
\[ x=0:\ \frac{y^2}{9}+\frac{z^2}{25}=1, \quad y=0:\ \frac{x^2}{4}+\frac{z^2}{25}=1 \]
Verify the axis intercepts
Why: Setting the other two variables to zero gives x = plus or minus 2, y = plus or minus 3, z = plus or minus 5, matching an ellipsoid with those semi-axes.
\[ (\pm 2,0,0),\ (0,\pm 3,0),\ (0,0,\pm 5) \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Trace an ellipsoid", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Setting the other two variables to zero gives x = plus or minus 2, y = plus or minus 3, z = plus or minus 5, matching an ellipsoid with those semi-axes.
Concept
Two variables are squared with the same sign; the third appears to the first power. It is a bowl opening along the linear-variable axis.
\[ z = \frac{x^2}{a^2} + \frac{y^2}{b^2} \]
Horizontal traces are ellipses that grow with height; vertical traces are parabolas. The vertex sits at the bottom of the bowl.
Concept
Two squared variables with opposite signs, and the third to the first power. This is the saddle, or Pringle chip, shape.
\[ z = \frac{x^2}{a^2} - \frac{y^2}{b^2} \]
Horizontal traces are hyperbolas; the two vertical traces are parabolas opening in opposite directions. The origin is a saddle point.
Concept
All three variables squared, one sign opposite the other two, equal to zero. The equals-zero is what makes it a cone rather than a hyperboloid.
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z^2}{c^2} \]
Horizontal traces are ellipses that shrink to a single point at the origin (the vertex); vertical traces through the axis are pairs of lines.
Concept
All three squared, one minus sign, equal to one. It is a single connected surface, pinched in the middle like a cooling tower.
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \]
Horizontal traces are ellipses at every height (never empty); vertical traces are hyperbolas. The axis is the variable with the minus sign.
Concept
All three squared, two minus signs, equal to one. It splits into two separate bowl-shaped pieces facing away from each other.
\[ -\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \]
There is a gap near the origin where no surface exists; the axis is the single variable with the plus sign.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Given the equation below, a student sees a subtraction and jumps to two sheets.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Guessing from the mere presence of a minus sign ignores how many minus signs there are.
Write the equation with the right side equal to one, then count the minus signs. The count equals the number of sheets.
Why: Guessing from the mere presence of a minus sign ignores how many minus signs there are. This surface is actually connected, not split.
Trap
Given the equation below, a student sees a subtraction and jumps to two sheets.
\[ x^2 + y^2 - z^2 = 1 \]
Wrong: calls it two sheets
Why: Guessing from the mere presence of a minus sign ignores how many minus signs there are. This surface is actually connected, not split.
\[ \text{claim: two sheets (incorrect)} \]
Write the equation with the right side equal to one, then count the minus signs. The count equals the number of sheets.
Right: one minus sign means one sheet
Why: There is exactly one negative term, so it is a hyperboloid of one sheet.
\[ \text{one minus sign} \Rightarrow \text{one sheet} \]
Confirm with the z = 0 trace
Why: At z = 0 the equation becomes x squared plus y squared equals 1, a real circle. Since the surface exists at the middle, it is connected -- one sheet.
\[ z=0:\ x^2 + y^2 = 1 \ \text{(a real circle, so connected)} \]
Notation
Annotate
From Trap: one sheet vs two sheets — read this one piece at a time. What is each part doing?
On: \( z=0:\ x^2 + y^2 = 1 \ \text{(a real circle, so connected)} \)
Worked example
Identify the surface below.
\[ \frac{x^2}{4} + \frac{y^2}{9} - z^2 = 1 \]
Count squared terms and minus signs
Why: Three squared terms, one minus sign, equal to one -- this points to a hyperboloid of one sheet.
Check horizontal traces
Why: Setting z to any constant leaves an ellipse with a positive right side, so a trace exists at every height.
\[ z=k:\ \frac{x^2}{4}+\frac{y^2}{9} = 1 + k^2 > 0 \]
Check a vertical trace
Why: Setting x to zero gives a hyperbola, matching the pinched-tower shape.
\[ x=0:\ \frac{y^2}{9} - z^2 = 1 \ \text{(hyperbola)} \]
Verify it is connected
Why: Because every horizontal trace is a nonempty ellipse, including at z = 0, the surface is a single connected piece, confirming a hyperboloid of one sheet.
\[ z=0:\ \frac{x^2}{4}+\frac{y^2}{9}=1 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Identify a quadric by its traces", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Setting z to any constant leaves an ellipse with a positive right side, so a trace exists at every height.
Pattern
1. Count the squared variables
Why: One squared missing (a linear term instead) points to a paraboloid; all three squared points to an ellipsoid, cone, or hyperboloid.
2. Read the sign pattern and the right side
Why: Equal to one with zero minus signs is an ellipsoid; one minus sign is one sheet; two minus signs is two sheets. Equal to zero with a mixed sign is a cone.
3. Confirm with traces
Why: Horizontal and vertical traces (ellipses, parabolas, hyperbolas) verify the classification and reveal the axis and orientation.
Check
Identify the surface described by the equation.
\[ z^2 - x^2 - y^2 = 1 \]
Check your understanding
Which quadric surface is this?
Answer: A
Why: Written with the right side equal to one, there are two minus signs (on x squared and y squared), so it is a hyperboloid of two sheets. Setting z = 0 gives -x^2 - y^2 = 1, which has no solution, so there is a gap and the surface splits into two pieces.
Concept
Cylindrical coordinates keep the height coordinate but describe the base using polar coordinates instead of x and y.
\[ (r, \theta, z) \]
Here r is the distance from the z-axis, theta is the angle around it, and z is the same height as before.
They fit any situation with symmetry around an axis: cylinders, cones, and paraboloids.
Intuition
Picture the flat polar system on the floor: how far out and which way around. Then just add how high up.
That third number, the height, is untouched from rectangular coordinates. Only the floor position changes to distance-and-angle.
Concept
The conversion is exactly the polar conversion in the floor plane, with the height carried straight across.
\[ x = r\cos\theta, \quad y = r\sin\theta, \quad z = z \]
Going the other way, distance and angle come from x and y:
\[ r^2 = x^2 + y^2, \quad \tan\theta = \frac{y}{x} \]
Estimation
Predict first
Convert the cylindrical point to rectangular coordinates.
Commit before you compute: what does Convert cylindrical to rectangular come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the radius
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The sum x squared plus y squared is 4 plus 12 = 16, whose square root is 4, matching the original radius, so the conversion checks out.
Worked example
Convert the cylindrical point to rectangular coordinates.
\[ (r, \theta, z) = \left(4, \tfrac{\pi}{3}, 5\right) \]
Compute the x-coordinate
Why: Multiply the radius by the cosine of the angle.
\[ x = 4\cos\tfrac{\pi}{3} = 4\cdot\tfrac{1}{2} = 2 \]
Compute the y-coordinate
Why: Multiply the radius by the sine of the angle.
\[ y = 4\sin\tfrac{\pi}{3} = 4\cdot\tfrac{\sqrt{3}}{2} = 2\sqrt{3} \]
Carry the height across
Why: The z-coordinate is unchanged between the two systems.
\[ z = 5, \qquad (x,y,z) = (2, 2\sqrt{3}, 5) \]
Verify the radius
Why: The sum x squared plus y squared is 4 plus 12 = 16, whose square root is 4, matching the original radius, so the conversion checks out.
\[ x^2 + y^2 = 4 + 12 = 16 = 4^2 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Convert cylindrical to rectangular", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sum x squared plus y squared is 4 plus 12 = 16, whose square root is 4, matching the original radius, so the conversion checks out.
Step zero
Discussion prompt
Convert rectangular to cylindrical — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the radius
Answer:
Worked example
Convert the rectangular point to cylindrical coordinates.
\[ (x, y, z) = (1, 1, 3) \]
Find the radius
Why: The radius is the distance from the z-axis, the square root of x squared plus y squared.
\[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \]
Find the angle
Why: Since x and y are both positive and equal, the point is in the first quadrant on the line y equals x.
\[ \tan\theta = \tfrac{1}{1} = 1 \ \Rightarrow\ \theta = \tfrac{\pi}{4} \]
Carry the height across
Why: The z-coordinate stays the same in cylindrical coordinates.
\[ (r,\theta,z) = \left(\sqrt{2}, \tfrac{\pi}{4}, 3\right) \]
Verify by converting back
Why: The radius times cosine of the angle is root two times root two over two, which equals 1, matching the original x, so the conversion is correct.
\[ r\cos\theta = \sqrt{2}\cdot\tfrac{\sqrt{2}}{2} = 1 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Convert rectangular to cylindrical", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The radius times cosine of the angle is root two times root two over two, which equals 1, matching the original x, so the conversion is correct.
Concept
Spherical coordinates locate a point by one distance and two angles.
\[ (\rho, \varphi, \theta) \]
Here rho is the straight-line distance from the origin, phi is the angle down from the positive z-axis, and theta is the same around-the-axis angle as in cylindrical.
The angle phi ranges from zero to pi (top to bottom), while theta ranges over a full turn.
\[ 0 \le \varphi \le \pi, \qquad 0 \le \theta \le 2\pi \]
Explain it
Discussion prompt
Explain Spherical coordinates to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Spherical coordinates locate a point by one distance and two angles.
Intuition
Think of a globe. Rho is how far the point is from the center. Phi is like latitude measured from the north pole down. Theta is like longitude around the equator.
The north pole is phi equal to zero; the equator is phi equal to a quarter turn; the south pole is phi equal to a half turn.
Analogy
Discussion prompt
Explain Distance plus latitude plus longitude by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of a globe. Rho is how far the point is from the center. Phi is like latitude measured from the north pole down. Theta is like longitude around the equator.
Concept
The height comes from the cosine of phi; the floor radius comes from the sine of phi, then splits into x and y by theta.
\[ x = \rho\sin\varphi\cos\theta, \quad y = \rho\sin\varphi\sin\theta, \quad z = \rho\cos\varphi \]
And the distance back from rectangular coordinates:
\[ \rho^2 = x^2 + y^2 + z^2 \]
Counterexample
Discussion prompt
The height comes from the cosine of phi; the floor radius comes from the sine of phi, then splits into x and y by theta.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Ranking
Put in order
Put the moves of Convert spherical to rectangular into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Use rho times sine of phi times cosine of theta.
Worked example
Convert the spherical point to rectangular coordinates.
\[ (\rho, \varphi, \theta) = \left(4, \tfrac{\pi}{3}, \tfrac{\pi}{6}\right) \]
Compute the x-coordinate
Why: Use rho times sine of phi times cosine of theta.
\[ x = 4\sin\tfrac{\pi}{3}\cos\tfrac{\pi}{6} = 4\cdot\tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt3}{2} = 3 \]
Compute the y-coordinate
Why: Use rho times sine of phi times sine of theta.
\[ y = 4\sin\tfrac{\pi}{3}\sin\tfrac{\pi}{6} = 4\cdot\tfrac{\sqrt3}{2}\cdot\tfrac{1}{2} = \sqrt{3} \]
Compute the z-coordinate
Why: The height is rho times cosine of phi.
\[ z = 4\cos\tfrac{\pi}{3} = 4\cdot\tfrac{1}{2} = 2, \quad (x,y,z)=(3,\sqrt{3},2) \]
Verify the distance
Why: The sum of squares is 9 plus 3 plus 4 = 16, whose square root is 4, matching the original rho, so the conversion is correct.
\[ x^2+y^2+z^2 = 9+3+4 = 16 = 4^2 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Convert spherical to rectangular", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sum of squares is 9 plus 3 plus 4 = 16, whose square root is 4, matching the original rho, so the conversion is correct.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A student treats phi as the around-the-axis angle and lets it run through a full turn, or plugs phi where theta belongs.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Letting phi go up to a full turn and using it for the horizontal angle double-counts points and puts them in the wrong place.
Keep the roles straight: phi is the tilt down from the positive z-axis and stops at a half turn; theta is the spin around the axis and goes a full turn.
Why: Letting phi go up to a full turn and using it for the horizontal angle double-counts points and puts them in the wrong place. Phi only measures down from the top.
Trap
A student treats phi as the around-the-axis angle and lets it run through a full turn, or plugs phi where theta belongs.
Wrong: uses phi as the azimuth
Why: Letting phi go up to a full turn and using it for the horizontal angle double-counts points and puts them in the wrong place. Phi only measures down from the top.
\[ 0 \le \varphi \le 2\pi \ \text{(wrong range)} \]
Keep the roles straight: phi is the tilt down from the positive z-axis and stops at a half turn; theta is the spin around the axis and goes a full turn.
Right: phi from the axis, theta around it
Why: Phi ranges only from zero to pi so it never repeats a direction; theta handles the full rotation. The z-coordinate always uses cosine of phi.
\[ 0 \le \varphi \le \pi, \quad 0 \le \theta \le 2\pi \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Pattern
1. Cylindrical is polar plus a height
Why: Use x = r cosine theta, y = r sine theta, and keep z. Reverse with r as the square root of x squared plus y squared.
2. Spherical uses one distance and two angles
Why: Use x = rho sine phi cosine theta, y = rho sine phi sine theta, z = rho cosine phi, with phi measured from the positive z-axis.
3. Always sanity-check the radius or distance
Why: Recomputing r squared or rho squared from the rectangular answer catches sign and angle mistakes immediately.
Real world
Discussion prompt
Outside this lesson: where does Week 2 - Geometry of Space actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Recipe: coordinate conversions is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers lines and planes in space and the distances between them, then cylinders, the six quadric surfaces, and the cylindrical and spherical coordinate systems. It targets the confusion between a normal vector and a direction vector, the sign pattern that separates a one-sheet surface from a two-sheet one, and the absolute-value error in the distance-to-a-plane formula.
Check
Convert the spherical point to rectangular coordinates.
\[ (\rho, \varphi, \theta) = \left(2, \tfrac{\pi}{2}, \pi\right) \]
Check your understanding
What are the rectangular coordinates?
Answer: A
Why: With phi = pi/2, sine of phi is 1 and cosine of phi is 0, so z = 0. Then x = 2(1)cos(pi) = -2 and y = 2(1)sin(pi) = 0, giving the point (-2, 0, 0).
Concept
The whole family sorts by squared terms, sign pattern, and right-hand side.
| Surface | Squared terms | Right side / signs |
|---|---|---|
| Ellipsoid | all three | equals one, all plus |
| Elliptic paraboloid | two (one linear) | same signs, linear term set equal |
| Hyperbolic paraboloid | two (one linear) | opposite signs, saddle |
| Cone | all three | equals zero, one sign opposite |
| Hyperboloid of one sheet | all three | equals one, one minus sign |
| Hyperboloid of two sheets | all three | equals one, two minus signs |
Discrimination
Sort into buckets
Sort these by Squared terms, from memory, without looking back at The quadric surfaces at a glance. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Recipe: writing a line's equations · Recipe: writing a plane's equation · Recipe: point-to-plane distance · Recipe: identify a quadric surface · Recipe: coordinate conversions. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A line is fixed by a point and a direction vector; a plane by a point and a normal vector. Keep those two roles distinct.
Angles between planes come from the angle between normals; distances to a plane use the absolute-value formula, and distance to a line uses a cross product over the direction's length.
Identify quadrics by squared terms and sign pattern -- count the minus signs to tell one sheet from two.
Cylindrical coordinates are polar plus a height; spherical coordinates use one distance and two angles, with the tilt angle measured from the positive z-axis.
| Object | What you need |
|---|---|
| Line | a point and a direction vector |
| Plane | a point and a normal vector |
| Distance to a plane | absolute value over normal length |
| Distance to a line | cross product over direction length |
| Quadric type | squared terms plus sign pattern |
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