This 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.
Subject: Calculus III · 110 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Tell a vector from a scalar and write a vector in component form in the plane and in space.
2. Use the space coordinate system: find distances and write the equation of a sphere.
3. Add, scale, and normalize vectors, and work with the standard basis.
4. Use the dot product to find angles, test orthogonality, and compute projections and work.
5. Use the cross product to build a perpendicular vector, find areas, and use the triple scalar product for volume.
Concept
A scalar is a single number that measures size only: a temperature, a mass, a speed.
A vector carries two facts at once: how much, and which way. Velocity, force, and displacement are vectors.
vector — A quantity with both magnitude (length) and direction, drawn as an arrow. Two arrows are the same vector if they have the same length and direction, no matter where they sit.
Counterexample
Discussion prompt
A scalar is a single number that measures size only: a temperature, a mass, a speed.
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.
Answer:
A vector carries two facts at once: how much, and which way. Velocity, force, and displacement are vectors.
Picture it
Figure (svg): Two identical arrows pointing up and to the right, drawn at different starting points, labeled as the same vector.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Think of giving directions: walk 5 blocks is a scalar, but walk 5 blocks north is a vector. The word north is the extra fact a scalar cannot hold.
Intuition
Think of giving directions: walk 5 blocks is a scalar, but walk 5 blocks north is a vector. The word north is the extra fact a scalar cannot hold.
Because only length and direction matter, you can slide an arrow anywhere in the plane without changing the vector it represents.
Figure (svg): Two identical arrows pointing up and to the right, drawn at different starting points, labeled as the same vector.
Analogy
Discussion prompt
Explain An arrow bundles two answers 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 giving directions: walk 5 blocks is a scalar, but walk 5 blocks north is a vector. The word north is the extra fact a scalar cannot hold.
Concept
To pin a vector to numbers, record how far it moves horizontally and vertically. These are its components.
\[ \mathbf{v} = \langle v_1, v_2 \rangle \]
For a vector from a start point to an end point, subtract: end minus start, coordinate by coordinate.
\[ \overrightarrow{PQ} = \langle q_1 - p_1,\ q_2 - p_2 \rangle \]
Explain it
Discussion prompt
Explain Component form in the plane 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:
To pin a vector to numbers, record how far it moves horizontally and vertically. These are its components.
Ranking
Put in order
Put the moves of Components from 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 components record the change in position, which is end minus start.
Worked example
Find the component form of the vector from the point P to the point Q.
\[ P = (-1, 2), \qquad Q = (3, 5) \]
Subtract start from end in each coordinate
Why: The components record the change in position, which is end minus start.
\[ \overrightarrow{PQ} = \langle 3 - (-1),\ 5 - 2 \rangle = \langle 4, 3 \rangle \]
State the answer
Why: This vector moves 4 to the right and 3 up.
\[ \overrightarrow{PQ} = \langle 4, 3 \rangle \]
Verify with the length
Why: The arrow should have a sensible length; four squared plus three squared is twenty-five, whose root is five.
\[ \lVert \overrightarrow{PQ} \rVert = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \]
Picture it
Animation
Shows: Each line of the worked example "Components from two points", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The arrow should have a sensible length; four squared plus three squared is twenty-five, whose root is five.
Concept
A position vector is the special arrow that runs from the origin out to a point. Its components are just the coordinates of that point.
\[ \text{point } (a, b) \;\longleftrightarrow\; \text{position vector } \langle a, b \rangle \]
This is why points and vectors look alike on paper: the position vector is the bridge between them.
Concept
Two vectors are equal exactly when their components match, one slot at a time. Location on the page is irrelevant.
\[ \langle a_1, a_2 \rangle = \langle b_1, b_2 \rangle \iff a_1 = b_1 \text{ and } a_2 = b_2 \]
So a single vector has many drawings but one component form.
Concept
Space needs three axes. We add a third axis, the vertical one, meeting the other two at right angles at the origin.
\[ \text{a point in space} = (x, y, z) \]
The three axes split space into eight regions called octants, just as two axes split the plane into four quadrants.
Picture it
Figure (svg): Three coordinate axes meeting at an origin: x toward the viewer, y to the right, z upward.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Point the fingers of your right hand along the first axis and curl them toward the second axis. Your thumb then points along the third axis. That is the standard orientation.
Intuition
Point the fingers of your right hand along the first axis and curl them toward the second axis. Your thumb then points along the third axis. That is the standard orientation.
Figure (svg): Three coordinate axes meeting at an origin: x toward the viewer, y to the right, z upward.
Concept
Everything from the plane carries over: a vector in space has three components, and a vector between points is still end minus start.
\[ \mathbf{v} = \langle v_1, v_2, v_3 \rangle, \qquad \overrightarrow{PQ} = \langle q_1 - p_1,\ q_2 - p_2,\ q_3 - p_3 \rangle \]
Concept
Distance between two points is the length of the vector connecting them. It is the Pythagorean theorem with one more term.
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Step zero
Discussion prompt
Distance between two space points — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Take the differences in each coordinate
Answer:
Worked example
Find the distance between the points A and B.
\[ A = (1, 1, 1), \qquad B = (3, 4, 7) \]
Take the differences in each coordinate
Why: These differences are the components of the connecting vector.
\[ \langle 3-1,\ 4-1,\ 7-1 \rangle = \langle 2, 3, 6 \rangle \]
Square, add, and take the root
Why: The length of a vector is the root of the sum of the squares of its components.
\[ d = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} \]
State the answer
Why: The root of forty-nine is exactly seven.
\[ d = 7 \]
Verify the arithmetic
Why: Rebuild the sum under the root: four plus nine is thirteen, plus thirty-six is forty-nine, and seven squared is forty-nine.
\[ 7^2 = 49 = 4 + 9 + 36 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Distance between two space points", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Rebuild the sum under the root: four plus nine is thirteen, plus thirty-six is forty-nine, and seven squared is forty-nine.
Concept
A sphere is the set of all points at a fixed distance from a center. Set the distance formula equal to the radius and square both sides.
\[ (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2 \]
Here the center is the point with coordinates a, b, c, and the radius is r.
Estimation
Predict first
An equation is given in expanded form. Find the center and radius.
Commit before you compute: what does Find the center and radius by completing the square 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 re-expanding the constants
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 added constants four, nine, and one total fourteen; twenty-five minus fourteen returns the original constant eleven.
Worked example
An equation is given in expanded form. Find the center and radius.
\[ x^2 + y^2 + z^2 - 4x + 6y - 2z - 11 = 0 \]
Group each variable and move the constant over
Why: Completing the square works on one variable group at a time.
\[ (x^2 - 4x) + (y^2 + 6y) + (z^2 - 2z) = 11 \]
Complete each square
Why: Add the square of half each linear coefficient to both sides: half of negative four is negative two, half of six is three, half of negative two is negative one.
\[ (x-2)^2 + (y+3)^2 + (z-1)^2 = 11 + 4 + 9 + 1 \]
Read off the center and radius
Why: The right side is twenty-five, so the radius is its root, five.
\[ (x-2)^2 + (y+3)^2 + (z-1)^2 = 25, \quad \text{center } (2,-3,1),\ r = 5 \]
Verify by re-expanding the constants
Why: The added constants four, nine, and one total fourteen; twenty-five minus fourteen returns the original constant eleven.
\[ 25 - (4 + 9 + 1) = 25 - 14 = 11 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Find the center and radius by completing the square", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The added constants four, nine, and one total fourteen; twenty-five minus fourteen returns the original constant eleven.
Pattern
1. To get a vector between points, subtract end minus start
Why: This works in the plane and in space, coordinate by coordinate.
2. For a length or distance, square each component, add, take the root
Why: Distance is just the length of the connecting vector.
3. For a sphere given expanded, complete the square in each variable
Why: This rewrites it as center-radius form so you can read both directly.
Elimination
Eliminate the wrong options
What is the distance from the origin to the point (2, 3, 6)?
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: Square each coordinate, add, and take the root: four plus nine plus thirty-six is forty-nine, and the root of forty-nine is seven.
Check
Find the distance from the origin to the given point.
\[ O = (0,0,0), \qquad P = (2, 3, 6) \]
Check your understanding
What is the distance from the origin to the point (2, 3, 6)?
Answer: A
Why: Square each coordinate, add, and take the root: four plus nine plus thirty-six is forty-nine, and the root of forty-nine is seven.
Concept
You add or subtract vectors slot by slot: add the first components, add the second components, and so on.
\[ \mathbf{u} + \mathbf{v} = \langle u_1 + v_1,\ u_2 + v_2,\ u_3 + v_3 \rangle \]
\[ \mathbf{u} - \mathbf{v} = \langle u_1 - v_1,\ u_2 - v_2,\ u_3 - v_3 \rangle \]
Picture it
Figure (svg): Two arrows placed tip to tail with a third arrow, the sum, drawn from the first tail to the second tip.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
To add arrows, place the tail of the second at the tip of the first. The sum runs from the very start to the very end.
Intuition
To add arrows, place the tail of the second at the tip of the first. The sum runs from the very start to the very end.
Figure (svg): Two arrows placed tip to tail with a third arrow, the sum, drawn from the first tail to the second tip.
Concept
Multiplying a vector by a number stretches or shrinks it and, if the number is negative, flips its direction.
\[ c\,\mathbf{v} = \langle c\,v_1,\ c\,v_2,\ c\,v_3 \rangle \]
The number is called a scalar precisely because it scales the vector.
Intuition
Doubling a vector gives an arrow twice as long in the same direction. Multiplying by a negative one keeps the length but reverses the arrow.
Any nonzero multiple keeps the vector on the same line through the origin; that is what parallel means.
Fill the middle
Fill in the blanks
From Compute a linear combination — finish the line. Write what belongs on the right of the equals sign before you look.
2\mathbf\langle 4-0,\ -2-12,\ 6-(-6) \rangle - 3\mathbf___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Distribute the number through every component before combining.
Worked example
Evaluate the combination of the two vectors.
\[ \mathbf{u} = \langle 2, -1, 3 \rangle, \quad \mathbf{v} = \langle 0, 4, -2 \rangle, \quad \text{find } 2\mathbf{u} - 3\mathbf{v} \]
Scale each vector first
Why: Distribute the number through every component before combining.
\[ 2\mathbf{u} = \langle 4, -2, 6 \rangle, \qquad 3\mathbf{v} = \langle 0, 12, -6 \rangle \]
Subtract component by component
Why: Subtraction of vectors happens one slot at a time.
\[ 2\mathbf{u} - 3\mathbf{v} = \langle 4-0,\ -2-12,\ 6-(-6) \rangle \]
State the answer
Why: Simplify each slot.
\[ 2\mathbf{u} - 3\mathbf{v} = \langle 4, -14, 12 \rangle \]
Verify the middle slot
Why: The second component is negative two minus twelve, which is negative fourteen, matching the answer.
\[ -2 - 12 = -14 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Compute a linear combination", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The second component is negative two minus twelve, which is negative fourteen, matching the answer.
Concept
The magnitude is the length of the arrow. It is a scalar and is never negative.
\[ \lVert \mathbf{v} \rVert = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
magnitude — The length of a vector, found as the square root of the sum of the squares of its components. Written with double bars around the vector.
Definition probe
Sort into buckets
Every line below is part of the definition of vector or of magnitude — one or the other, never both. Put each where it belongs.
Fill the middle
Fill in the blanks
From Magnitude in space — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\langle 2, -3, 6 \rangle = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Squaring removes the sign, so direction does not affect length.
Worked example
Find the length of the vector.
\[ \mathbf{v} = \langle 2, -3, 6 \rangle \]
Square each component
Why: Squaring removes the sign, so direction does not affect length.
\[ 2^2 + (-3)^2 + 6^2 = 4 + 9 + 36 \]
Add and take the root
Why: The sum under the root is forty-nine.
\[ \lVert \mathbf{v} \rVert = \sqrt{49} = 7 \]
Verify it is positive and sensible
Why: A length must be nonnegative, and seven exceeds the largest component six, as a diagonal should.
\[ 7 > 6 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Magnitude in space", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A length must be nonnegative, and seven exceeds the largest component six, as a diagonal should.
Concept
A unit vector has length one. It records pure direction with no size.
To normalize a vector, divide it by its own magnitude.
\[ \hat{\mathbf{u}} = \frac{\mathbf{v}}{\lVert \mathbf{v} \rVert} \]
normalize — To scale a nonzero vector to length one by dividing each component by the vector's magnitude. The result points the same way.
Step zero
Discussion prompt
Find the unit vector — 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 magnitude first
Answer:
Worked example
Find the unit vector in the direction of the vector below.
\[ \mathbf{v} = \langle 2, -3, 6 \rangle \]
Find the magnitude first
Why: You must know the length before you can divide by it.
\[ \lVert \mathbf{v} \rVert = \sqrt{4 + 9 + 36} = 7 \]
Divide every component by the magnitude
Why: Dividing by the length rescales the arrow to length one without turning it.
\[ \hat{\mathbf{u}} = \left\langle \tfrac{2}{7},\ -\tfrac{3}{7},\ \tfrac{6}{7} \right\rangle \]
Verify the length is one
Why: Squaring and adding the components gives forty-nine over forty-nine, whose root is one.
\[ \sqrt{\tfrac{4}{49} + \tfrac{9}{49} + \tfrac{36}{49}} = \sqrt{\tfrac{49}{49}} = 1 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Find the unit vector", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Squaring and adding the components gives forty-nine over forty-nine, whose root is one.
Trap
Asked for the unit vector in the direction of the vector below, a student just reports the vector itself.
\[ \mathbf{v} = \langle 2, -3, 6 \rangle \;\Rightarrow\; \text{claim } \hat{\mathbf{u}} = \langle 2, -3, 6 \rangle \]
But this arrow has length seven, not one, so it is not a unit vector.
\[ \lVert \langle 2,-3,6 \rangle \rVert = 7 \ne 1 \]
Divide by the magnitude so the length becomes one.
\[ \hat{\mathbf{u}} = \frac{1}{7}\langle 2, -3, 6 \rangle = \left\langle \tfrac{2}{7},\ -\tfrac{3}{7},\ \tfrac{6}{7} \right\rangle \]
Now the length checks out as one.
\[ \left\lVert \hat{\mathbf{u}} \right\rVert = \frac{7}{7} = 1 \checkmark \]
Concept
Three special unit vectors point along the three axes. Every vector is a combination of them.
\[ \mathbf{i} = \langle 1,0,0 \rangle,\quad \mathbf{j} = \langle 0,1,0 \rangle,\quad \mathbf{k} = \langle 0,0,1 \rangle \]
\[ \langle a, b, c \rangle = a\,\mathbf{i} + b\,\mathbf{j} + c\,\mathbf{k} \]
Estimation
Predict first
Find the vector of length ten that points in the direction of the vector a.
Commit before you compute: what does A vector of a given length in a given direction 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 length is ten
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. Six squared plus eight squared is one hundred, whose root is ten.
Worked example
Find the vector of length ten that points in the direction of the vector a.
\[ \mathbf{a} = \langle 3, 0, 4 \rangle \]
Normalize a to get pure direction
Why: The magnitude of a is the root of nine plus sixteen, which is five.
\[ \hat{\mathbf{a}} = \frac{1}{5}\langle 3, 0, 4 \rangle = \left\langle \tfrac{3}{5}, 0, \tfrac{4}{5} \right\rangle \]
Scale the unit vector to length ten
Why: A unit vector times ten has length ten and keeps the direction.
\[ 10\,\hat{\mathbf{a}} = \langle 6, 0, 8 \rangle \]
Verify the length is ten
Why: Six squared plus eight squared is one hundred, whose root is ten.
\[ \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A vector of a given length in a given direction", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Six squared plus eight squared is one hundred, whose root is ten.
Pattern
Combine: scale each vector, then add or subtract slot by slot
Why: Every linear combination reduces to these two moves.
Length: square the components, add, take the root
Why: This is the magnitude, always a nonnegative scalar.
Direction: divide the vector by its magnitude to normalize
Why: This produces the unit vector; multiply that by any length you want.
Check
Find the magnitude of the vector below.
\[ \mathbf{v} = \langle 1, -4, 8 \rangle \]
Check your understanding
What is the magnitude of the vector with components 1, negative 4, and 8?
Answer: A
Why: One squared plus four squared plus eight squared is one plus sixteen plus sixty-four, which is eighty-one, and the root of eighty-one is nine.
Concept
The dot product multiplies two vectors and gives back a single scalar, not a vector.
\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \]
Multiply matching components, then add the results. The output is one number.
Intuition
The dot product is large and positive when two vectors point the same way, zero when they are perpendicular, and negative when they point apart.
So its sign alone tells you whether the angle between the vectors is sharp, square, or wide.
Concept
There is a geometric form of the dot product that brings in the angle between the two vectors.
\[ \mathbf{u} \cdot \mathbf{v} = \lVert \mathbf{u} \rVert\, \lVert \mathbf{v} \rVert \cos\theta \]
Solve for the cosine to recover the angle from the components.
\[ \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\lVert \mathbf{u} \rVert\, \lVert \mathbf{v} \rVert} \]
Ranking
Put in order
Put the moves of Angle between two vectors 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 angle formula divides by the product of the lengths.
Worked example
Find the angle between the two vectors.
\[ \mathbf{u} = \langle 1, 2, 2 \rangle, \qquad \mathbf{v} = \langle 4, 0, 3 \rangle \]
Compute the dot product
Why: Multiply matching components and add.
\[ \mathbf{u} \cdot \mathbf{v} = (1)(4) + (2)(0) + (2)(3) = 10 \]
Compute both magnitudes
Why: The angle formula divides by the product of the lengths.
\[ \lVert \mathbf{u} \rVert = \sqrt{1+4+4} = 3, \qquad \lVert \mathbf{v} \rVert = \sqrt{16+0+9} = 5 \]
Divide to get the cosine, then invert
Why: The cosine is the dot product over the product of the lengths.
\[ \cos\theta = \frac{10}{(3)(5)} = \frac{2}{3}, \qquad \theta = \arccos\tfrac{2}{3} \approx 48.2^\circ \]
Verify the cosine is in range
Why: Two-thirds lies between negative one and one, so it is a valid cosine of a real angle.
\[ -1 \le \tfrac{2}{3} \le 1 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Angle between two vectors", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Two-thirds lies between negative one and one, so it is a valid cosine of a real angle.
Concept
Because the magnitudes are always positive, the sign of the dot product is exactly the sign of the cosine of the angle.
| dot product | angle is | vectors |
|---|---|---|
| positive | acute | point roughly together |
| zero | right angle | perpendicular |
| negative | obtuse | point roughly apart |
Comparison
Comparison matrix
From The sign of the dot product: refill the angle is column from what you know. The rest of the table is as it appeared.
| dot product | angle is | vectors |
|---|---|---|
| positive | acute | point roughly together |
| zero | right angle | perpendicular |
| negative | obtuse | point roughly apart |
Concept
Two nonzero vectors are orthogonal, meaning perpendicular, exactly when their dot product is zero.
\[ \mathbf{u} \perp \mathbf{v} \iff \mathbf{u} \cdot \mathbf{v} = 0 \]
This is the fastest perpendicularity test there is: one multiply-and-add.
Worked example
Find the value of k that makes the two vectors orthogonal.
\[ \mathbf{u} = \langle 2, k, 3 \rangle, \qquad \mathbf{v} = \langle 4, -2, k \rangle \]
Set the dot product equal to zero
Why: Perpendicular vectors have a dot product of zero.
\[ (2)(4) + (k)(-2) + (3)(k) = 0 \]
Simplify and solve
Why: Combine the k terms and isolate k.
\[ 8 - 2k + 3k = 0 \;\Rightarrow\; 8 + k = 0 \;\Rightarrow\; k = -8 \]
Verify by substituting back
Why: With k equal to negative eight the dot product is eight plus sixteen minus twenty-four, which is zero.
\[ (2)(4) + (-8)(-2) + (3)(-8) = 8 + 16 - 24 = 0 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Solve for orthogonality", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: With k equal to negative eight the dot product is eight plus sixteen minus twenty-four, which is zero.
Trap
A student is asked for a scalar measuring alignment but reaches for the cross product, which is meant for a perpendicular vector.
\[ \text{asked: how aligned? } \Rightarrow \text{wrongly compute } \mathbf{u} \times \mathbf{v} \]
The cross product returns a vector, so it cannot be the number the angle formula needs.
\[ \mathbf{u} \times \mathbf{v} = \text{a vector, not a scalar} \]
Alignment and angle come from the dot product, which returns a scalar.
\[ \mathbf{u} \cdot \mathbf{v} = \lVert \mathbf{u} \rVert\, \lVert \mathbf{v} \rVert \cos\theta \;\; (\text{a number}) \]
Rule of thumb: dot gives a number for angles and projections; cross gives a vector perpendicular to both.
\[ \text{dot} \to \text{scalar}, \qquad \text{cross} \to \text{vector} \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Alignment and angle come from the dot product, which returns a scalar.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Pattern
Angle: dot product over the product of the two lengths, then take the inverse cosine
Why: This inverts the geometric form of the dot product.
Perpendicular test: check whether the dot product equals zero
Why: A zero dot product means a right angle.
Sign read: positive is acute, zero is right, negative is obtuse
Why: The magnitudes are positive, so the sign is the cosine's sign.
Prediction
Predict first
What is the angle between the vectors 1,0,1 and 1,1,0?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: 60 degrees
Why: The dot product is one, each magnitude is the root of two, so the cosine is one over two, and the inverse cosine of one half is sixty degrees.
Check
Find the angle between the two vectors.
\[ \mathbf{u} = \langle 1, 0, 1 \rangle, \qquad \mathbf{v} = \langle 1, 1, 0 \rangle \]
Check your understanding
What is the angle between the vectors 1,0,1 and 1,1,0?
Answer: A
Why: The dot product is one, each magnitude is the root of two, so the cosine is one over two, and the inverse cosine of one half is sixty degrees.
Concept
The scalar projection of one vector onto another is how much of the first lies along the second. It is a single number.
\[ \operatorname{comp}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\lVert \mathbf{a} \rVert} \]
Notice you divide by the length of a just once.
Concept
The vector projection points the scalar amount back along the direction of a. It is a vector.
\[ \operatorname{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{\lVert \mathbf{a} \rVert^2}\, \mathbf{a} \]
Here you divide by the length squared and then attach the vector a, so the result carries a direction.
Picture it
Figure (svg): Vector b above a horizontal vector a, with a dashed drop line showing b's shadow on the line of a.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Shine a light straight down onto the line carrying a. The shadow of b on that line is the projection.
Intuition
Shine a light straight down onto the line carrying a. The shadow of b on that line is the projection.
The length of the shadow is the scalar projection; the shadow itself, sitting on the line, is the vector projection.
Figure (svg): Vector b above a horizontal vector a, with a dashed drop line showing b's shadow on the line of a.
Step zero
Discussion prompt
Scalar and vector projection — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Compute the dot product and the length of a
Answer:
Worked example
Find both the scalar and the vector projection of b onto a.
\[ \mathbf{a} = \langle 2, 2, 1 \rangle, \qquad \mathbf{b} = \langle 4, 1, 2 \rangle \]
Compute the dot product and the length of a
Why: Both projection formulas need these two pieces.
\[ \mathbf{a} \cdot \mathbf{b} = 8 + 2 + 2 = 12, \qquad \lVert \mathbf{a} \rVert = \sqrt{4+4+1} = 3 \]
Scalar projection: divide by the length once
Why: The scalar projection uses the first power of the length of a.
\[ \operatorname{comp}_{\mathbf{a}} \mathbf{b} = \frac{12}{3} = 4 \]
Vector projection: divide by the length squared and attach a
Why: The length squared is nine, and the result must be a vector along a.
\[ \operatorname{proj}_{\mathbf{a}} \mathbf{b} = \frac{12}{9}\langle 2,2,1 \rangle = \left\langle \tfrac{8}{3}, \tfrac{8}{3}, \tfrac{4}{3} \right\rangle \]
Verify the vector projection's length equals the scalar projection
Why: The vector projection is four-thirds times a, whose length is three, giving length four, matching the scalar projection.
\[ \tfrac{12}{9}\,\lVert \mathbf{a} \rVert = \tfrac{4}{3}\cdot 3 = 4 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Scalar and vector projection", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The vector projection is four-thirds times a, whose length is three, giving length four, matching the scalar projection.
Trap
Asked for the vector projection, a student stops at the scalar projection and reports a lone number.
\[ \text{report } 4 \text{ as the vector projection} \]
A number has no direction, so it cannot be the vector answer the question wants.
\[ 4 \ne \left\langle \tfrac{8}{3}, \tfrac{8}{3}, \tfrac{4}{3} \right\rangle \]
Read the question. If it says scalar or component, give the number; if it says vector, attach the direction a.
\[ \operatorname{comp}_{\mathbf{a}}\mathbf{b} = 4, \qquad \operatorname{proj}_{\mathbf{a}}\mathbf{b} = \left\langle \tfrac{8}{3}, \tfrac{8}{3}, \tfrac{4}{3} \right\rangle \]
The scalar projection divides by the length once; the vector projection divides by the length squared and multiplies by a.
Concept
When a constant force pushes an object along a straight displacement, the work done is the dot product of force and displacement.
\[ W = \mathbf{F} \cdot \mathbf{d} \]
Only the part of the force along the motion does work, which is exactly what the dot product measures.
Hypothesis
Predict first
Work done by a constant force 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: Build the displacement vector
Why: Displacement is end minus start, here just the coordinates of Q.
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
A constant force moves an object from the origin to the point Q. Find the work done.
\[ \mathbf{F} = \langle 2, 3, 4 \rangle, \qquad O = (0,0,0),\ Q = (1, 2, 2) \]
Build the displacement vector
Why: Displacement is end minus start, here just the coordinates of Q.
\[ \mathbf{d} = \langle 1, 2, 2 \rangle \]
Dot the force with the displacement
Why: Work is the dot product of force and displacement.
\[ W = (2)(1) + (3)(2) + (4)(2) = 2 + 6 + 8 = 16 \]
Verify the sign is sensible
Why: The force and displacement point generally the same way, so the work is positive, which matches sixteen.
\[ W = 16 > 0 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Work done by a constant force", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The force and displacement point generally the same way, so the work is positive, which matches sixteen.
Check
Find the scalar projection of b onto a.
\[ \mathbf{a} = \langle 2, -1, 2 \rangle, \qquad \mathbf{b} = \langle 4, 1, 4 \rangle \]
Check your understanding
What is the scalar projection of b onto a?
Answer: A
Why: The dot product is eight minus one plus eight, which is fifteen, and the length of a is three, so the scalar projection is fifteen over three, equal to five.
Concept
The cross product takes two vectors in space and produces a third vector perpendicular to both. It lives only in three dimensions.
\[ \mathbf{a} \times \mathbf{b} \perp \mathbf{a} \quad\text{and}\quad \mathbf{a} \times \mathbf{b} \perp \mathbf{b} \]
This is its whole reason for existing: it manufactures a direction perpendicular to a plane.
Concept
Write a symbolic determinant with the basis vectors on top, the components of a in the middle, and the components of b on the bottom.
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
Expanding along the top row gives each component. Watch the minus sign on the middle term.
\[ \mathbf{a} \times \mathbf{b} = \langle a_2 b_3 - a_3 b_2,\ -(a_1 b_3 - a_3 b_1),\ a_1 b_2 - a_2 b_1 \rangle \]
Fill the middle
Fill in the blanks
From Compute a cross product — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\langle -3, 6, -3 \rangle \times \mathbf___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Cover the first column: two times six minus three times five.
Worked example
Find the cross product of a and b.
\[ \mathbf{a} = \langle 1, 2, 3 \rangle, \qquad \mathbf{b} = \langle 4, 5, 6 \rangle \]
First component: the i minor
Why: Cover the first column: two times six minus three times five.
\[ (2)(6) - (3)(5) = 12 - 15 = -3 \]
Second component: the j minor with a sign flip
Why: The middle term carries a leading minus: negative of one times six minus three times four.
\[ -\big[(1)(6) - (3)(4)\big] = -(6 - 12) = 6 \]
Third component: the k minor
Why: Cover the third column: one times five minus two times four.
\[ (1)(5) - (2)(4) = 5 - 8 = -3 \]
Assemble the result
Why: Collect the three components.
\[ \mathbf{a} \times \mathbf{b} = \langle -3, 6, -3 \rangle \]
Verify it is perpendicular to a
Why: Dotting the result with a must give zero if the cross product is correct.
\[ \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = -3 + 12 - 9 = 0 \checkmark \]
Concept
Two directions are perpendicular to a plane. The right-hand rule picks which one the cross product chooses.
Point your right fingers along a and curl them toward b. Your thumb points along the cross product.
Figure (svg): Vectors a and b in a horizontal plane with the cross product pointing straight up out of the plane.
Explain it
Discussion prompt
Explain Direction: the right-hand rule 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:
Two directions are perpendicular to a plane. The right-hand rule picks which one the cross product chooses.
Trap
A student assumes order does not matter and swaps the factors freely, as with ordinary multiplication.
\[ \text{assume } \mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{a} \]
But swapping the order reverses the direction, so the two answers are opposites.
\[ \mathbf{b} \times \mathbf{a} = \langle 3, -6, 3 \rangle \ne \langle -3, 6, -3 \rangle \]
Swapping the order flips the sign of the whole vector.
\[ \mathbf{a} \times \mathbf{b} = -\,(\mathbf{b} \times \mathbf{a}) \]
Keep the factors in the order the problem gives them; the right-hand rule reverses if you flip them.
\[ \mathbf{a} \times \mathbf{a} = \mathbf{0} \]
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.
walk 5 blocks is a scalar, but walk 5 blocks north is a vector. The word north is the extra fact a scalar cannot hold.; To pin a vector to numbers, record how far it moves horizontally and vertically. These are its components.Concept
The length of the cross product equals the area of the parallelogram spanned by the two vectors.
\[ \lVert \mathbf{a} \times \mathbf{b} \rVert = \lVert \mathbf{a} \rVert\, \lVert \mathbf{b} \rVert \sin\theta = \text{area of the parallelogram} \]
Half of that is the area of the triangle with the same two edges.
Analogy
Discussion prompt
Explain Magnitude of the cross product is an area 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:
The length of the cross product equals the area of the parallelogram spanned by the two vectors.
Estimation
Predict first
Find the area of the triangle with the three given vertices.
Commit before you compute: what does Area of a triangle 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 with base and height
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. Edge AC has length three along a horizontal line, and B sits two units above it, giving one half times three times two, which is three.
Worked example
Find the area of the triangle with the three given vertices.
\[ A = (1,1,0),\ B = (2,3,0),\ C = (4,1,0) \]
Build two edge vectors from A
Why: The triangle is spanned by the edges leaving one common vertex.
\[ \overrightarrow{AB} = \langle 1, 2, 0 \rangle, \qquad \overrightarrow{AC} = \langle 3, 0, 0 \rangle \]
Cross the two edges
Why: The magnitude of the cross product is the parallelogram area.
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \langle 0, 0, -6 \rangle \]
Take half the magnitude
Why: A triangle is half of the parallelogram; the length here is six.
\[ \text{area} = \tfrac{1}{2}\lVert \langle 0,0,-6 \rangle \rVert = \tfrac{1}{2}(6) = 3 \]
Verify with base and height
Why: Edge AC has length three along a horizontal line, and B sits two units above it, giving one half times three times two, which is three.
\[ \tfrac{1}{2}(3)(2) = 3 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Area of a triangle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Edge AC has length three along a horizontal line, and B sits two units above it, giving one half times three times two, which is three.
Step zero
Discussion prompt
A unit vector perpendicular to two vectors — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Cross the vectors to get a perpendicular direction
Answer:
Worked example
Find a unit vector perpendicular to both a and b.
\[ \mathbf{a} = \langle 1, 2, 3 \rangle, \qquad \mathbf{b} = \langle 4, 5, 6 \rangle \]
Cross the vectors to get a perpendicular direction
Why: The cross product is automatically perpendicular to both inputs.
\[ \mathbf{a} \times \mathbf{b} = \langle -3, 6, -3 \rangle \]
Find its magnitude
Why: You must normalize the perpendicular vector to get length one.
\[ \lVert \mathbf{a} \times \mathbf{b} \rVert = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6} \]
Divide to normalize
Why: Dividing by the magnitude produces a unit vector in the same direction.
\[ \hat{\mathbf{n}} = \frac{1}{3\sqrt{6}}\langle -3, 6, -3 \rangle = \frac{1}{\sqrt{6}}\langle -1, 2, -1 \rangle \]
Verify it is perpendicular and unit length
Why: Its dot with a is negative one plus four minus three, which is zero, and its component squares sum to six over six, giving length one.
\[ \mathbf{a} \cdot \hat{\mathbf{n}} = \tfrac{-1+4-3}{\sqrt{6}} = 0, \quad \lVert \hat{\mathbf{n}} \rVert = \sqrt{\tfrac{6}{6}} = 1 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "A unit vector perpendicular to two vectors", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Its dot with a is negative one plus four minus three, which is zero, and its component squares sum to six over six, giving length one.
Pattern
Compute: expand the determinant, remembering the minus on the middle term
Why: This produces the perpendicular vector component by component.
Perpendicular direction: normalize the cross product
Why: The cross product is perpendicular to both inputs; divide by its length for a unit normal.
Area: the magnitude is the parallelogram area, half of it the triangle area
Why: The sine factor makes the magnitude an area rather than an alignment.
Real world
Discussion prompt
Outside this lesson: where does Week 1 - Vectors & Their Properties 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 The cross-product toolkit 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 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.
Elimination
Eliminate the wrong options
What is a cross b for a = 2,1,0 and b = 1,3,0?
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: Both vectors lie in the horizontal plane, so the i and j components are zero, and the k component is two times three minus one times one, which is five.
Check
Compute the cross product of a and b.
\[ \mathbf{a} = \langle 2, 1, 0 \rangle, \qquad \mathbf{b} = \langle 1, 3, 0 \rangle \]
Check your understanding
What is a cross b for a = 2,1,0 and b = 1,3,0?
Answer: A
Why: Both vectors lie in the horizontal plane, so the i and j components are zero, and the k component is two times three minus one times one, which is five.
Concept
Dotting one vector with the cross product of two others gives a single number called the triple scalar product.
\[ \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} \]
Its absolute value is the volume of the parallelepiped built on the three vectors.
\[ V = \big| \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \big| \]
Counterexample
Discussion prompt
Dotting one vector with the cross product of two others gives a single number called the triple scalar product.
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.
Answer:
Its absolute value is the volume of the parallelepiped built on the three vectors.
Ranking
Put in order
Put the moves of Volume of a parallelepiped 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 triple scalar product is this determinant of the three edge vectors.
Worked example
Find the volume of the parallelepiped with edges u, v, and w.
\[ \mathbf{u} = \langle 1,1,0 \rangle,\ \mathbf{v} = \langle 0,1,1 \rangle,\ \mathbf{w} = \langle 1,0,1 \rangle \]
Set up the three-by-three determinant
Why: The triple scalar product is this determinant of the three edge vectors.
\[ \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} \]
Expand along the top row
Why: Use cofactor expansion with the alternating sign pattern.
\[ 1(1\cdot 1 - 1\cdot 0) - 1(0\cdot 1 - 1\cdot 1) + 0 = 1(1) - 1(-1) = 2 \]
Take the absolute value for the volume
Why: Volume cannot be negative, so take the absolute value of the triple scalar product.
\[ V = |2| = 2 \]
Verify the sign handling
Why: The determinant came out positive, so the absolute value leaves it unchanged at two.
\[ |2| = 2 \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Volume of a parallelepiped", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The determinant came out positive, so the absolute value leaves it unchanged at two.
Concept
Ask what kind of answer the problem wants. A number about angle, alignment, projection, or work calls for the dot product.
A vector perpendicular to a plane, or an area or volume, calls for the cross product or the triple scalar product.
| You want | Use | Result |
|---|---|---|
| angle or projection | dot product | scalar |
| work | dot product | scalar |
| perpendicular vector | cross product | vector |
| parallelogram area | cross product | scalar |
| parallelepiped volume | triple scalar product | scalar |
Discrimination
Sort into buckets
Sort these by Use, from memory, without looking back at Dot or cross: which do I want?. 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 — Setting up points, vectors, and spheres · The vector-operation toolkit · The dot-product toolkit · The cross-product toolkit · Vectors carry direction; scalars do not. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can write vectors in component form in the plane and in space, and move between points and position vectors.
You can find distances and sphere equations, and you can add, scale, measure, and normalize vectors.
You can use the dot product for angles, orthogonality, projections, and work, always getting a scalar.
You can use the cross product for a perpendicular vector and for areas, and the triple scalar product for volume.
Keep the two products straight: dot gives a number, cross gives a vector, and the cross product reverses sign when you swap its factors.
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