MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: cross product, areas and volumes (MATH 133)

This sheet is not a summary of Nicholson 4.3: you already have the course notes. It answers one question only, what makes students lose marks on the cross product, areas and volumes in MATH 133 at McGill University, and which precise gesture avoids each loss.

Everything below is done by hand, as on the exam: small integer vectors, exact square roots, no calculator. The one habit that pays across the whole chapter is the ten-second check $u \cdot (u \times v) = v \cdot (u \times v) = 0$ , written on the copy right after the product, before anything is built on it.

The thread of the chapter

The cross product answers with a VECTOR, and that vector has two halves: its direction is a normal, its length is an area. Every question of the chapter uses exactly one half, and the marks are lost by using the wrong half, or by building $u \times v$ from the wrong pair of vectors.

This chapter is part of MATH 133, Linear Algebra and Geometry (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Cross product
Right-hand rule
Lagrange identity
Area of a parallelogram
Triangle area in space
Scalar triple product
Volume of a parallelepiped
Volume of a tetrahedron
Coplanarity of four points
Normal vector of a plane
Point-to-line distance
Distance between skew lines

The essentials

One vector, two halves

• Definition: $u \times v = (u_2v_3 - u_3v_2,\ u_3v_1 - u_1v_3,\ u_1v_2 - u_2v_1)$ , the first-row expansion of the symbolic determinant with $i, j, k$ on top. The $j$ term carries a MINUS sign.
• DIRECTION: perpendicular to $u$ and to $v$ , oriented by the right-hand rule. So $v \times u = -(u \times v)$ , and $u \times v$ is a normal to any plane containing $u$ and $v$ .
• LENGTH: $\|u \times v\| = \|u\|\|v\|\sin\theta$ , the area of the parallelogram on $u$ and $v$ . Independent check by Lagrange: $\|u \times v\|^2 = \|u\|^2\|v\|^2 - (u \cdot v)^2$ .
• $u \times v = 0$ exactly when $u$ and $v$ are parallel: the parallelogram has collapsed, and there is no normal to read.

The same

u \times v

answers two questions: its direction is the normal to the plane of

u

and

v

, its length is the area of their parallelogram.

v \times u

has the same length and points the other way.

Before computing, write on the copy which half the question needs: “normal” and “direction of the line” use the direction, “area”, “volume” and “distance” use the length. Half of the lost marks in this chapter are a correct vector given where a number was asked, or the reverse.

What each classic question feeds to the product

• Triangle $ABC$ : area $\frac{1}{2}\|\overrightarrow{AB} \times \overrightarrow{AC}\|$ , both edges leaving the SAME vertex.
• Plane through $A$ , $B$ , $C$ : $n = \overrightarrow{AB} \times \overrightarrow{AC}$ , then $n \cdot (X - A) = 0$ . Plane through a line $(A, d)$ and a point $P$ : $n = d \times \overrightarrow{AP}$ .
• Volume: parallelepiped $|u \cdot (v \times w)|$ , tetrahedron $\frac{1}{6}$ of it; four points coplanar exactly when $\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = 0$ .
• Line of two planes: direction $n_1 \times n_2$ . Distance from $Q$ to the line $(P, d)$ : $\frac{\|\overrightarrow{PQ} \times d\|}{\|d\|}$ . Skew lines: $\frac{|\overrightarrow{P_1P_2} \cdot (d_1 \times d_2)|}{\|d_1 \times d_2\|}$ .

Every one of these formulas is area divided by base or volume divided by base area. If you remember the picture, you can rebuild the formula in the exam room.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The algebra of $u \times v$ : what survives from ordinary products

Distributivity and scalars survive; commutativity, associativity and cancellation do not. A red box is not an answer: it is the order to rewrite the expression.

Expression	Tempting rewrite	What is true
$v \times u$	$u \times v$	$-(u \times v)$
Example: $(0, 1, 0) \times (1, 0, 0) = (0, 0, -1)$ , the opposite of $(1, 0, 0) \times (0, 1, 0) = (0, 0, 1)$ .
$u \times u$	$\\|u\\|^{2}$	$0$
Example: $(3, 1, 2) \times (3, 1, 2) = (0, 0, 0)$ , a vector, not the number $14$ .
$u \times (v + w)$	$u \times v + u \times w$	$u \times v + u \times w$
Example: $(1, 0, 0) \times (0, 1, 1) = (0, -1, 1) = (0, 0, 1) + (0, -1, 0)$ .
$(u \times v) \times w$	$u \times (v \times w)$	no regrouping rule that does not exist
Example: $(i \times i) \times j = (0, 0, 0)$ but $i \times (i \times j) = i \times k = (0, -1, 0)$ . What to do: Compute the inner bracket first, in the order written, and never move the parentheses.
$u \times v = u \times w$	$v = w$	$v - w$ parallel to $u$ rule that does not exist
Example: $(1, 0, 0) \times (0, 1, 0) = (1, 0, 0) \times (1, 1, 0) = (0, 0, 1)$ with $(0, 1, 0) \neq (1, 1, 0)$ . What to do: Write $u \times (v - w) = 0$ and conclude only that $v - w = cu$ for some scalar $c$ .

The last two rows are where the real-number reflexes break. Both are tested on midterms as true or false statements.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Dropping the minus sign on the middle component

the whole question, and every plane or distance built on the vector

What not to write

“ $(3, 1, 2) \times (1, -2, 4) = (1 \cdot 4 - 2 \cdot (-2),\ 3 \cdot 4 - 2 \cdot 1,\ 3 \cdot (-2) - 1 \cdot 1) = (8, 10, -7)$ .”

What to write

“The $j$ minor enters with a minus sign: $(8, -10, -7)$ . Check: $(3, 1, 2) \cdot (8, -10, -7) = 0$ and $(1, -2, 4) \cdot (8, -10, -7) = 0$ .”

Why: The wrong vector gives $(3, 1, 2) \cdot (8, 10, -7) = 20 \neq 0$ : it is not even perpendicular to $u$ . The two dot products take ten seconds and catch this slip every time, which is why they belong on the copy and not in your head.

2. Forgetting the half in the area of a triangle

1 mark, on a question where everything else was right

What not to write

“ $\overrightarrow{AB} \times \overrightarrow{AC} = (-6, -2, 6)$ , so the triangle has area $\sqrt{76} = 2\sqrt{19}$ .”

What to write

“ $\sqrt{76} = 2\sqrt{19}$ is the area of the parallelogram on $\overrightarrow{AB}$ and $\overrightarrow{AC}$ , so triangle $ABC$ has area $\frac{1}{2} \times 2\sqrt{19} = \sqrt{19}$ .”

Why: The cross product measures the PARALLELOGRAM on the two vectors, and the diagonal $BC$ cuts it into two copies of the triangle. Here $A = (1, 1, 0)$ , $B = (3, 1, 2)$ , $C = (1, 4, 1)$ . Say “parallelogram” in words before halving, and the half cannot disappear.

3. Using position vectors instead of edge vectors

the whole question

What not to write

“Area of $ABC = \frac{1}{2}\|\overrightarrow{OA} \times \overrightarrow{OC}\| = \frac{1}{2}\|(1, -1, 3)\| = \frac{\sqrt{11}}{2}$ .”

What to write

“Area of $ABC = \frac{1}{2}\|\overrightarrow{AB} \times \overrightarrow{AC}\| = \frac{1}{2}\|(-6, -2, 6)\| = \sqrt{19}$ : both edges leave from $A$ .”

The triangle asked for is

ABC

; the dashed pair

\overrightarrow{OA}

\overrightarrow{OC}

builds the triangle

OAC

instead, whose area changes as soon as the origin moves.

Why: $\overrightarrow{OA} \times \overrightarrow{OC}$ is the area of the triangle $OAC$ , a different triangle with a vertex at the origin. An area cannot depend on where the origin is; position vectors do. Differences of points are the only vectors that belong to the figure.

4. Patching a zero cross product into a plane equation

the whole question, plus a false equation

What not to write

“ $\overrightarrow{AB} \times \overrightarrow{AC} = (0, 0, 0)$ , so the plane through $A$ , $B$ , $C$ is $0x + 0y + 0z = 0$ .”

What to write

“ $\overrightarrow{AC} = 3\overrightarrow{AB}$ , so the cross product is zero and the three points are collinear: infinitely many planes contain them, and the question has no unique answer.”

Why: With $A = (2, 0, 1)$ , $B = (3, 2, 2)$ , $C = (5, 6, 4)$ : $\overrightarrow{AB} = (1, 2, 1)$ and $\overrightarrow{AC} = (3, 6, 3)$ . The zero vector has no direction, so it is not a normal, and $0 = 0$ describes all of space. A zero cross product is information about the points, never an arithmetic accident.

5. Plane through a line and a point: crossing d with OP instead of AP

the whole question

What not to write

“ $L: (0, 1, 1) + t(1, 0, 2)$ and $P = (1, 1, 0)$ . $n = d \times \overrightarrow{OP} = (-2, 2, 1)$ , so the plane is $-2x + 2y + z = 0$ .”

What to write

“ $A = (0, 1, 1)$ is on $L$ and $\overrightarrow{AP} = (1, 0, -1)$ , so $n = d \times \overrightarrow{AP} = (0, 3, 0)$ and the plane is $y = 1$ .”

Why: Both vectors fed to the product must lie IN the plane. $\overrightarrow{OP}$ starts at the origin, which is not on $L$ . The wrong plane even misses the line: $A$ gives $0 + 2 + 1 = 3 \neq 0$ . Check $n \cdot d = 0$ and one point of $L$ before writing the final answer.

6. One third instead of one sixth for a tetrahedron

2 marks, and a volume twice too large

What not to write

“The tetrahedron on $(2, 0, 0)$ , $(0, 3, 0)$ , $(0, 0, 4)$ is a pyramid, so $V = \frac{1}{3} \times 24 = 8$ .”

What to write

“ $V = \frac{1}{6}|u \cdot (v \times w)| = \frac{24}{6} = 4$ . Check: base triangle $\frac{1}{2} \times 2 \times 3 = 3$ , height $4$ , $\frac{1}{3} \times 3 \times 4 = 4$ .”

Why: A pyramid is one third of the prism on the same base, but the base of a tetrahedron is a TRIANGLE, half of the parallelogram that the triple product uses. One third times one half is one sixth.

7. Stopping at the area when the question asks for a distance

2 marks, and an answer wrong by the factor $\|d\|$

What not to write

“The distance from $Q = (3, 0, 0)$ to the line $t(1, 2, 2)$ is $\|\overrightarrow{OQ} \times d\| = \|(0, -6, 6)\| = 6\sqrt{2}$ .”

What to write

“ $\text{dist} = \dfrac{\|\overrightarrow{OQ} \times d\|}{\|d\|} = \dfrac{6\sqrt{2}}{3} = 2\sqrt{2}$ .”

Why: $\|\overrightarrow{PQ} \times d\|$ is the AREA of the parallelogram on $\overrightarrow{PQ}$ and $d$ ; the distance is its HEIGHT, so divide by the base $\|d\|$ . The quick test: the distance to a line can never exceed $\|\overrightarrow{PQ}\| = 3$ , and $6\sqrt{2} \approx 8.5$ does.

8. Setting to zero a variable that is constant along the line

the whole question, and a false conclusion

What not to write

“For $x + y + z = 3$ and $x - y + z = 1$ , set $y = 0$ : $x + z = 3$ and $x + z = 1$ . No solution, so the planes do not meet.”

What to write

“ $n_1 \times n_2 = (2, 0, -2)$ , so $d = (1, 0, -1)$ and $y$ is constant on the line: $y = 1$ . Set $z = 0$ instead: the point is $(2, 1, 0)$ .”

Why: Subtracting the equations gives $2y = 2$ , so every point of the line has $y = 1$ and none has $y = 0$ . The direction vector warns you in advance: never set to zero a variable whose component in $d$ is $0$ .

Which method to choose

Which half of the cross product the question needs

Read the NOUN the question asks for, before choosing any formula

The box on

u

v

w

has volume

|u \cdot (v \times w)|

; the tetrahedron cut from its corner by the three orange edges keeps one sixth of it, never one third.

If a normal, the equation of a plane, or the direction of a line of intersection → the DIRECTION: $\overrightarrow{AB} \times \overrightarrow{AC}$ , $d \times \overrightarrow{AP}$ or $n_1 \times n_2$ , then a point

Example: $(1, 2, 0)$ , $(3, 0, 1)$ , $(0, 1, 2)$ give $n = (3, 5, 4)$ and $3x + 5y + 4z = 13$
If the area of a parallelogram or of a triangle → the LENGTH, halved for a triangle, with edges from one vertex

Example: $\|(5, -5, 5)\| = 5\sqrt{3}$ , so the triangle has area $\frac{5\sqrt{3}}{2}$
If a volume, or whether four points are coplanar → the triple product $u \cdot (v \times w)$ , in absolute value, divided by $6$ for a tetrahedron

Example: $(1, 1, 0) \cdot \big((0, 2, 1) \times (1, 0, 3)\big) = 7$ : box $7$ , tetrahedron $\frac{7}{6}$
If the distance from a point to a line → the length divided by the base: $\frac{\|\overrightarrow{PQ} \times d\|}{\|d\|}$

Example: $\frac{\|(9, -6, -6)\|}{3} = \sqrt{17}$
If the distance between two non-parallel lines → the triple product divided by $\|d_1 \times d_2\|$ ; zero means the lines meet

Example: $\frac{|(1, 0, 5) \cdot (1, -1, 1)|}{\sqrt{3}} = 2\sqrt{3}$
If a torque about a given axis → the full vector $\tau = r \times F$ , then its component along the axis

Example: $(0.3, 0.4, 0) \times (0, 0, -50) = (-20, 15, 0)$ : length $25$ , turning part $0$

If a branch asks for a number and you are holding a vector, you are not finished; if it asks for a direction and you are holding a number, you took the length too early. The distance from a point to a PLANE is not on this tree: it uses the dot product with the normal, from the previous chapter.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

The plane through three points, as a marker wants to read it

When to use it: Find an equation of the plane through A, B and C, or through a line and a point

1 Compute two edge vectors from the SAME point: $\overrightarrow{AB}$ and $\overrightarrow{AC}$ (for a line and a point: $d$ and $\overrightarrow{AP}$ with $A$ on the line).
2 Compute their cross product with the three minors shown, and check it with two dot products equal to $0$ .
3 Say why it is not zero: “the points are not collinear”. If it IS zero, stop and say that no unique plane exists.
4 Write $n \cdot (X - A) = 0$ , expand, and simplify the coefficients by a common factor if there is one.
5 Check the equation with the two points that were NOT used for the constant.

Concluding sentence

“Since $\overrightarrow{AB} \times \overrightarrow{AC} = (6, 3, 2) \neq 0$ , the points are not collinear, and the plane through $A$ with normal $(6, 3, 2)$ is $6x + 3y + 2z = 6$ .”

The trap: Checking the equation only with the point used to build it. That point satisfies it by construction, so the check proves nothing; the other two points are the real test.

Marking: Typically 1 mark for the edge vectors, 2 for the cross product with its minors, 1 for the equation and 1 for the check. The cross product alone, correct and checked, is already worth more than a final equation with no working.

The distance between two skew lines

When to use it: Two lines given in parametric form, and a question about their distance or their position

1 Compute $n = d_1 \times d_2$ . If $n = 0$ the lines are parallel: use the point-to-line distance instead.
2 Compute $\overrightarrow{P_1P_2}$ between the two given points and the triple product $\overrightarrow{P_1P_2} \cdot n$ .
3 If it is $0$ , the lines are coplanar and not parallel, so they meet: the distance is $0$ . Otherwise they are skew.
4 Divide: $\text{dist} = \dfrac{|\overrightarrow{P_1P_2} \cdot n|}{\|n\|}$ , and simplify the root.

Concluding sentence

“ $d_1 \times d_2 = (0, -1, 1) \neq 0$ and $\overrightarrow{P_1P_2} \cdot (d_1 \times d_2) = 1 \neq 0$ , so the lines are skew and their distance is $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .”

The trap: Answering with $\|\overrightarrow{P_1P_2}\|$ , the distance between two particular points, which is only an upper bound.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The two dot products

Right after any cross product, dot it with both factors. Both results must be $0$ ; anything else means a minor or a sign is wrong.

$(3, 1, 2) \cdot (8, 10, -7) = 20$ : the middle sign is wrong. $(3, 1, 2) \cdot (8, -10, -7) = 0$ : fine.
Plug the points back in

A plane through three points must be satisfied by all three, and a plane through a line must also satisfy $n \cdot d = 0$ .

$6x + 3y + 2z = 6$ at $(0, 2, 0)$ gives $6$ and at $(0, 0, 3)$ gives $6$ : correct.
The size of a distance

The distance from $Q$ to a line can never exceed the distance from $Q$ to any point of that line. If it does, you forgot to divide by $\|d\|$ .

$\sqrt{17} \approx 4.1$ against $\|\overrightarrow{PQ}\| = \sqrt{21} \approx 4.6$ : plausible. $3\sqrt{17} \approx 12.4$ : impossible.
Six times a tetrahedron is an integer

When all four vertices have integer coordinates, the triple product is an integer, so $6V$ must be an integer too.

$V = \frac{8}{3}$ gives $6V = 16$ : fine. $V = \frac{2}{5}$ would give $6V = 2.4$ : wrong on sight.
The direction of an intersection line

The direction of $\Pi_1 \cap \Pi_2$ must be perpendicular to both normals, and your point must satisfy both equations.

$d = (0, 1, 1)$ with $n_1 = (1, -1, 1)$ and $n_2 = (2, 1, -1)$ : $0 - 1 + 1 = 0$ and $0 + 1 - 1 = 0$ .

The typical problem, taken apart

A tetrahedron: one face, its plane, the volume and a height

Let $A = (1, 0, 0)$ , $B = (0, 2, 0)$ , $C = (0, 0, 3)$ and $D = (2, 2, 2)$ .

Find the area of the face $ABC$ , the equation of its plane, the volume of the tetrahedron $ABCD$ , and the height from $D$ to the face $ABC$ .

Three vertices sit on the axes, so the face

ABC

is easy to picture; the height from

D

is measured perpendicular to that face, along the normal, not along any axis.

Step 1

$\overrightarrow{AB} = (-1, 2, 0)$ , $\overrightarrow{AC} = (-1, 0, 3)$ , and $\overrightarrow{AB} \times \overrightarrow{AC} = \big((2)(3) - (0)(0),\ (0)(-1) - (-1)(3),\ (-1)(0) - (2)(-1)\big) = (6, 3, 2)$ . Check: $-6 + 6 + 0 = 0$ and $-6 + 0 + 6 = 0$ .

Why

Both edges leave from $A$ , and the vector is checked before anything is built on it. This single product will serve three times: its length for the area, its direction for the plane, and inside the triple product for the volume.

Step 2

$\|(6, 3, 2)\| = \sqrt{36 + 9 + 4} = 7$ , so the face has area $\frac{7}{2}$ .

Why

The LENGTH is used here, and halved because the face is a triangle. The numbers were chosen so the root is exact, which is typical of MATH 133.

Step 3

Plane: $6(x - 1) + 3y + 2z = 0$ , that is $6x + 3y + 2z = 6$ . Check: $B$ gives $6$ , $C$ gives $6$ .

Why

The DIRECTION is used here, with the point $A$ . The check uses $B$ and $C$ , the points that did not build the constant.

Step 4

$\overrightarrow{AD} = (1, 2, 2)$ and $\overrightarrow{AD} \cdot (6, 3, 2) = 6 + 6 + 4 = 16$ , so $V = \frac{16}{6} = \frac{8}{3}$ .

Why

The triple product reuses the cross product already computed. One sixth, not one third: the base is a triangle. And $6V = 16$ is an integer, as it must be.

Step 5

Height from $D$ : $V = \frac{1}{3} \times \text{area} \times h$ gives $h = \frac{3V}{\text{area}} = \frac{8}{7/2} = \frac{16}{7}$ .

Why

The height is recovered from the volume without any new product: volume divided by base, once more.

Step 6

Check with the point-to-plane distance: $\dfrac{|6(2) + 3(2) + 2(2) - 6|}{7} = \dfrac{16}{7}$ .

Why

An independent route from the previous chapter gives the same number, so the plane, the volume and the area are all confirmed at once.

The conclusion, written out

“The face $ABC$ has area $\frac{7}{2}$ and lies in the plane $6x + 3y + 2z = 6$ ; the tetrahedron has volume $\frac{8}{3}$ , and the height from $D$ is $\frac{16}{7}$ .”

The classic mistake on this problem: Taking one third of the triple product: $V = \frac{16}{3}$ , and then $h = \frac{32}{7}$ , which the point-to-plane check immediately contradicts.

Learn by heart

• $u \times v = (u_2v_3 - u_3v_2,\ u_3v_1 - u_1v_3,\ u_1v_2 - u_2v_1)$ , then check $u \cdot (u \times v) = v \cdot (u \times v) = 0$ .
• $v \times u = -(u \times v)$ , $u \times u = 0$ , and NO associativity, NO cancellation.
• $\|u \times v\| = \|u\|\|v\|\sin\theta =$ area of the parallelogram; triangle: HALF of it.
• $|u \cdot (v \times w)| =$ volume of the box; tetrahedron: ONE SIXTH; zero means coplanar.
• Plane: $n = \overrightarrow{AB} \times \overrightarrow{AC}$ ; line of two planes: $d = n_1 \times n_2$ .
• Point to line: $\frac{\|\overrightarrow{PQ} \times d\|}{\|d\|}$ ; skew lines: $\frac{|\overrightarrow{P_1P_2} \cdot (d_1 \times d_2)|}{\|d_1 \times d_2\|}$ .
• Direction or length: decide which half the question needs BEFORE computing.

Frequently asked questions

How do I remember the formula for the cross product?

Do not memorise the components, memorise the determinant. Put i, j and k on the first row, u on the second and v on the third, and expand along the first row with the signs plus, minus, plus. The minus in front of the j term is the only trap. Then check the result by taking its dot product with u and with v: both must be zero.

How do I find the equation of a plane through three points?

Take two edge vectors from the same point, for example from A to B and from A to C, and compute their cross product: that is a normal vector. If it is the zero vector, the points are collinear and no unique plane exists. Otherwise write normal dot (X minus A) equals zero, expand, and check the equation with the two other points.

Why is the volume of a tetrahedron one sixth of the triple product?

The triple product gives the volume of the parallelepiped built on the three edges. A tetrahedron is a pyramid, one third of the prism on the same base, and its base is a triangle, which is half of the parallelogram used by the box. One third times one half gives one sixth. The corner of the unit cube, with volume one sixth, is the example to remember.

How do I find the distance from a point to a line in 3D?

Take a point P on the line, its direction d, and the vector from P to your point Q. The length of the cross product of PQ with d is the area of a parallelogram whose base is the length of d and whose height is the distance you want. So divide that length by the length of d. Forgetting the division is the classic mistake.

What does it mean when the cross product is zero?

It means the two vectors are parallel, or one of them is zero: the parallelogram they span has no area. In a plane question it means the three points are collinear, so no unique plane passes through them. For two planes it means their normals are parallel, so the planes are parallel or identical and there is no line of intersection to find.

Practise it

Corrected exercises: Cross product, areas and volumes, MATH 133 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: cross product, areas and volumes (MATH 133)

The essentials

One vector, two halves

What each classic question feeds to the product

The rules in table form

The algebra of u×vu \times vu×v: what survives from ordinary products

The mistakes that cost marks

1. Dropping the minus sign on the middle component

2. Forgetting the half in the area of a triangle

3. Using position vectors instead of edge vectors

4. Patching a zero cross product into a plane equation

5. Plane through a line and a point: crossing d with OP instead of AP

6. One third instead of one sixth for a tetrahedron

7. Stopping at the area when the question asks for a distance

8. Setting to zero a variable that is constant along the line

Which method to choose

Which half of the cross product the question needs

How the answer is expected to be written

The plane through three points, as a marker wants to read it

The distance between two skew lines

Check before you hand in

The typical problem, taken apart

A tetrahedron: one face, its plane, the volume and a height

Learn by heart

Frequently asked questions

How do I remember the formula for the cross product?

How do I find the equation of a plane through three points?

Why is the volume of a tetrahedron one sixth of the triple product?

How do I find the distance from a point to a line in 3D?

What does it mean when the cross product is zero?

Corrected exercises: Cross product, areas and volumes, MATH 133 at McGill

See also

Looking for a MATH 133 tutor in Montreal?

The algebra of $u \times v$ : what survives from ordinary products