On the Vectors page, a vector was an arrow: two numbers for the plane, three for space, and you could add arrows head to tail and stretch them. Nothing there told you how long an arrow is, or what angle two arrows make, and staring at the numbers won't tell you either. This page adds two new ways to multiply vectors that answer exactly those questions. The dot product takes two arrows and hands back a number, and that number knows about lengths and angles. The cross product takes two arrows in space and hands back a third arrow standing straight up off both of them, like a flagpole off a floor.
On Vector Spaces you could add vectors in $\R^n$ and stretch them, and that was it. Nothing there tells you how long a vector is, or what angle two vectors make. This page adds two new ways to multiply vectors that answer exactly those questions. The dot product takes two vectors and hands back a number, and that number knows about lengths and angles. The cross product only lives in $\R^3$, takes two vectors, and hands back a third vector standing perpendicular to both of them.
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away. Flip the switch on a question to type your answer instead: vectors as entries like (2, -1, 4), square roots like 2sqrt(5), and angles in degrees like 60.
On the precalculus Vectors page, the dot product of two arrows in the plane was $u_1v_1 + u_2v_2$, written u · v with little arrows over the letters. Space just adds a third pair to the sum. Same dot, plain letters, no arrows.
On the precalculus Vectors page, the dot product of two arrows in the plane was $u_1v_1 + u_2v_2$, written u · v with little arrows over the letters. Same idea here, with as many entries as you like, and with linear algebra's outfit: the same dot, but plain letters with no arrows.
Start with the simplest thing you could possibly do with two lists of numbers: line them up, multiply the matching entries, and add everything up.
The dot product of $u = (u_1, \ldots, u_n)$ and $v = (v_1, \ldots, v_n)$ in $\R^n$ is the number $$u \cdot v = u_1v_1 + u_2v_2 + \cdots + u_nv_n.$$
The dot product of $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$ in space is the number $$u \cdot v = u_1v_1 + u_2v_2 + u_3v_3.$$ In the plane, where vectors have two entries, drop the third term: $u \cdot v = u_1v_1 + u_2v_2$.
So the answer is always a single number, never a vector. Two quick ones, one in $\R^3$ and one in $\R^4$:
So the answer is always a single number, never a vector. Here's a way to see that number: each product $u_iv_i$ is the area of a rectangle with sides $u_i$ and $v_i$, counted positive when the two entries have the same sign and negative when they don't. The dot product is the total. Two quick ones:
Find (a) $(2, -1, 3) \cdot (4, 5, -2)$ (b) $(3, -1, 2) \cdot (1, 5, 1)$.
(a) $2 \cdot 4 + (-1) \cdot 5 + 3 \cdot (-2) = 8 - 5 - 6 = -3$.
(b) $3 \cdot 1 + (-1) \cdot 5 + 2 \cdot 1 = 3 - 5 + 2 = 0$. A zero dot product is going to mean something special. Hold that thought.
Both, as rectangles: red ones count positive, blue ones negative, and laid head to tail on a number line they land on the dot product. In (b) they cancel exactly.
Find (a) $(2, -1, 3) \cdot (4, 5, -2)$ (b) $(1, 0, -2, 3) \cdot (5, 7, 1, -1)$.
(a) $2 \cdot 4 + (-1) \cdot 5 + 3 \cdot (-2) = 8 - 5 - 6 = -3$.
(b) $1 \cdot 5 + 0 \cdot 7 + (-2) \cdot 1 + 3 \cdot (-1) = 5 + 0 - 2 - 3 = 0$. A zero dot product is going to mean something special. Hold that thought.
Before we use it, let's see which rules of ordinary multiplication survive. Most of them do.
For all $u, v, w$ in $\R^n$ and all real numbers $a, b$:
(1) $u_iv_i = v_iu_i$ for each $i$, so the two sums match term by term. (2) The $i$th entry of $au + bv$ is $au_i + bv_i$, so $(au + bv) \cdot w = \sum_i (au_i + bv_i)w_i = a\sum_i u_iw_i + b\sum_i v_iw_i$. (3) $u \cdot u = u_1^2 + \cdots + u_n^2$ is a sum of squares, so it's never negative, and it's $0$ only if every $u_i$ is $0$. $\blacksquare$
For all vectors $u, v, w$ in the plane or in space and all real numbers $a, b$:
This is a visual argument, not a formal proof. Draw $u \cdot v$ as its rectangles, one per entry, and watch what each rule does to them:
(1) Swapping $u$ and $v$ turns each rectangle on its side. A $2$ by $4$ rectangle and a $4$ by $2$ rectangle have the same area, so the total doesn't change. (2) Doubling $u$ doubles every rectangle's width, so it doubles every area and the total. Adding $v$ to $u$ glues a strip onto each rectangle, and the strips are exactly the rectangles of $v \cdot w$, so their areas add on. (3) Dot $u$ with itself and every rectangle becomes a square, $u_i$ by $u_i$. A square's area is never negative, so the total isn't either, and the only way to get $0$ is for every square to have side $0$, that is, $u = 0$. The full proof is in the n version.
Rule 2 (with rule 1, it works in either slot) is the one that saves time: $(3u - 2v) \cdot w$ is $3(u \cdot w) - 2(v \cdot w)$, so if you already know those two dot products, you're done. Rule 3 is the one that's about to give us lengths.
$u \cdot v$ is a number, so $(u \cdot v)w$ is a vector (that number times $w$), but $(u \cdot v) \cdot w$ is nonsense: you can't dot a number with a vector. And the classic slip is multiplying matching entries and then forgetting to add, which leaves you holding a vector. (Some books write the dot product as ⟨u, v⟩, or as uᵀv when vectors are columns. Same number.)
Your turn. Some of these build a combination first.
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Dot a vector with itself and you get $u_1^2 + u_2^2 + \cdots + u_n^2$$u_1^2 + u_2^2 + u_3^2$. In the plane, the square root of $u_1^2 + u_2^2$ is the length of the arrow, by Pythagoras. In space it's Pythagoras twice: once across the floor, once straight up.
The floor diagonal of the box is $\sqrt{2^2 + 3^2} = \sqrt{13}$, and the arrow is the hypotenuse of a second right triangle with legs $\sqrt{13}$ and $6$. So its length is $\sqrt{13 + 36} = 7$. That's exactly $\sqrt{v \cdot v}$, and it's how we define length in every dimensionin the plane and in space.
The length (or norm) of $u$ in $\R^n$ is $$\|u\| = \sqrt{u \cdot u} = \sqrt{u_1^2 + u_2^2 + \cdots + u_n^2}.$$
The length (or norm) of $u = (u_1, u_2, u_3)$ is $$\|u\| = \sqrt{u \cdot u} = \sqrt{u_1^2 + u_2^2 + u_3^2},$$ the length of its arrow. In the plane, $\|u\| = \sqrt{u_1^2 + u_2^2}$.
The distance between two points $p$ and $q$ is the length of the arrow from one to the other, $\|q - p\|$. And dividing a nonzero vector by its length, $\frac{u}{\|u\|}$, gives a unit vector pointing the same way: scaling by $c$ multiplies the length by $|c|$, since $\|cu\| = \sqrt{c^2(u \cdot u)} = |c| \, \|u\|$.
(a) Find $\|(1, -2, 2)\|$. (b) Find the distance between $p = (1, 0, 2)$ and $q = (4, -2, 8)$. (c) Find the unit vector pointing the same way as $(1, -2, 2)$.
(a) $\sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
(b) Subtract first: $q - p = (3, -2, 6)$, so the distance is $\sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
(c) Divide by the length from (a): $\frac{1}{3}(1, -2, 2) = \left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right)$. Check: $\frac{1}{9} + \frac{4}{9} + \frac{4}{9} = 1$.
Parts (b) and (c) in space: the arrow from $p$ to $q$ is the diagonal of a $3$ by $2$ by $6$ box, and dividing by $3$ shrinks $(1, -2, 2)$ down onto the ball of radius $1$.
Those whole-number answers are rigged, by the way: $(1, 2, 2)$, $(2, 3, 6)$, and $(4, 4, 7)$ are the space versions of the $3$-$4$-$5$ triangle. Most vectors give square roots, so simplify them all the way.
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Here's the surprise. The dot product doesn't just know lengths, it knows angles. Two vectors $u$ and $v$ always lie in one flat plane through $0$ (the span of the two)(lay a sheet of paper through the origin and both tips), so they make an honest angle $\theta$ between $0^\circ$ and $180^\circ$. Watch the dot product while one vector swings around the other:
Positive while the angle is acute, zero at exactly $90^\circ$, negative once it's obtuse. And it isn't just the sign that tracks the angle. The exact relationship is the same one you met in the plane:
If $\theta$ is the angle between the nonzero vectors $u$ and $v$, then $$u \cdot v = \|u\| \, \|v\| \cos\theta, \qquad \text{that is,} \qquad \cos\theta = \frac{u \cdot v}{\|u\| \, \|v\|}.$$
Draw $u$ and $v$ from $0$, inside the plane they span. The arrow $u - v$ runs from the tip of $v$ to the tip of $u$, so the three make a triangle with sides $\|u\|$, $\|v\|$, $\|u - v\|$ and the angle $\theta$ between the first two. The Law of Cosines says $$\|u - v\|^2 = \|u\|^2 + \|v\|^2 - 2\|u\| \, \|v\| \cos\theta.$$ Now expand the left side with the properties above: $$\|u - v\|^2 = (u - v) \cdot (u - v) = u \cdot u - 2(u \cdot v) + v \cdot v = \|u\|^2 - 2(u \cdot v) + \|v\|^2.$$ Comparing the two, $-2(u \cdot v) = -2\|u\| \, \|v\| \cos\theta$. (If $u$ and $v$ point the same way or opposite ways, the triangle is flat, and the Law of Cosines still holds with $\theta = 0^\circ$ or $180^\circ$.) $\blacksquare$
This is a visual argument, not a formal proof, but it's the real reason. Put $u$ and $v$ tail to tail. The arrow $u - v$ runs from $v$'s tip to $u$'s tip, so the three make a triangle, flat on one sheet of paper even in space. Draw a square on each side and swing $v$:
At exactly $90^\circ$ it's Pythagoras: the big square equals the two small ones. Close the angle and the far side shrinks, so the big square comes up short; open it and the big square overshoots. The rules above say by how much: $\|u - v\|^2 = (u - v) \cdot (u - v) = \|u\|^2 + \|v\|^2 - 2(u \cdot v)$, so the gap is exactly $2(u \cdot v)$. Trigonometry measures the same gap with the angle instead (that's the Law of Cosines): $2\|u\| \, \|v\| \cos\theta$. Same gap, so $u \cdot v = \|u\| \, \|v\| \cos\theta$. The full proof is in the n version.
So a zero dot product means a right angle. That's worth a word of its own: $u$ and $v$ are orthogonal when $u \cdot v = 0$. (In $\R^n$ with $n$ bigger than $3$ there's no picture to measure, so the formula becomes the definition of the angle. The Cauchy–Schwarz inequality is what guarantees that the fraction always lands between $-1$ and $1$.)In the plane and in space that's a real right angle, the kind you could check with a protractor.
Find the angle between (a) $u = (1, 1, 0)$ and $v = (0, 1, 1)$ (b) $u = (2, -1, 1)$ and $v = (1, 3, 1)$ (c) $u = (1, 2, -1)$ and $v = (-2, -1, -1)$.
(a) $u \cdot v = 0 + 1 + 0 = 1$ and both lengths are $\sqrt{2}$, so $\cos\theta = \frac{1}{2}$ and $\theta = 60^\circ$.
(b) $u \cdot v = 2 - 3 + 1 = 0$, so $\theta = 90^\circ$ with no lengths needed. They're orthogonal.
(c) $u \cdot v = -2 - 2 + 1 = -3$ and both lengths are $\sqrt{6}$, so $\cos\theta = -\frac{3}{6} = -\frac{1}{2}$ and $\theta = 120^\circ$. Negative dot product, obtuse angle, just like the animation said.
All three pairs in space, with their angles opening:
Now steer a vector yourself. Each mission asks for a different angle with the fixed vector $u$:
That first mission has lots of answers: every vector in a whole plane through $0$ is orthogonal to $u$. Keep that plane in mind, it's the star of the next page.
The zero vector is orthogonal to everything, since $u \cdot 0 = 0$, even though it makes no angle with anything. And watch the sign: the formula hands you a cosine, and a negative cosine means an obtuse angle. If you drop the sign, you'll land on the supplement ($60^\circ$ instead of $120^\circ$).
Your turn. The vectors are chosen so the cosine is a unit-circle value.
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Shine a light straight down onto the line through $u$, and $v$ casts a shadow on it. That shadow is a multiple of $u$, and finding the multiple is one dot product:
The shadow is the multiple $cu$ that makes the leftover $v - cu$ orthogonal to $u$. Setting $(v - cu) \cdot u = 0$ and solving gives $c = \frac{v \cdot u}{u \cdot u}$.
For $u \ne 0$, the projection of $v$ onto $u$ is $$\operatorname{proj}_u v = \frac{v \cdot u}{u \cdot u} \, u.$$
The word "orthogonal" in that derivation was the whole point, so it gets a theorem: the leftover really is orthogonal to $u$, and $v$ splits cleanly into a piece along $u$ and a piece across it.
For $u \ne 0$, the vector $v - \operatorname{proj}_u v$ is orthogonal to $u$. So $$v = \operatorname{proj}_u v + (v - \operatorname{proj}_u v)$$ splits $v$ into a multiple of $u$ plus a vector orthogonal to $u$.
Write $c = \frac{v \cdot u}{u \cdot u}$. By the properties of the dot product, $(v - cu) \cdot u = v \cdot u - c(u \cdot u) = v \cdot u - v \cdot u = 0$. $\blacksquare$
This is a visual argument, not a formal proof. Slide a candidate shadow $cu$ along the line and watch the gap arrow $v - cu$, from its tip to $v$'s tip. The numbers are the ones from the example below, $u = (1, 2, 2)$ and $v = (3, 0, 3)$:
Each time $c$ goes up by one, $(v - cu) \cdot u = v \cdot u - c(u \cdot u)$ drops by the same amount, $u \cdot u$. So it's a straight line (the little graph), it crosses zero exactly once, and that's exactly where the gap stands at a right angle. Solving $v \cdot u - c(u \cdot u) = 0$ gives $c = \frac{v \cdot u}{u \cdot u}$, the number in the definition. The leftover is perpendicular because we picked $c$ to make it so. The full proof is in the n version.
The shadow is also the point on $u$'s line closest to the tip of $v$. Try to beat it:
Every other point is the hypotenuse of a right triangle whose leg is the shortest gap, so nothing beats the foot of the perpendicular. That fact is how the next page finds the distance from a point to a plane. One by hand first.
Let $u = (1, 2, 2)$ and $v = (3, 0, 3)$. Find $\operatorname{proj}_u v$ and the leftover $v - \operatorname{proj}_u v$.
$v \cdot u = 3 + 0 + 6 = 9$ and $u \cdot u = 1 + 4 + 4 = 9$, so $\operatorname{proj}_u v = \frac{9}{9}u = (1, 2, 2)$. The leftover is $(3, 0, 3) - (1, 2, 2) = (2, -2, 1)$. Check: $(2, -2, 1) \cdot u = 2 - 4 + 2 = 0$.
That split in space: $v$ is the diagonal of a rectangle standing on edge, one side along $u$ and the other square to it.
Your turn. Divide by $u \cdot u$, the length squared.
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On Inner Products & Norms, one picture from this page becomes a famous inequality: a shadow is never longer than the arrow that casts it, so $|u \cdot v| \le \|u\| \, \|v\|$. And its cousin: a detour is never shorter than the straight path.
Everything so far used only the three properties of the dot product. On Inner Products & Norms, any product with those properties gets to be called an inner product, and lengths, angles, and projections all come along for free.
The dot product gave us a number. In $\R^3$ (and only there) there's a product that gives back a vector, and it's built to stand perpendicular to the two you started with. The formula looks random at first. It isn't, and the next theorem shows why.
The cross product of $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$ in $\R^3$ is the vector $$u \times v = (u_2v_3 - u_3v_2, \ u_3v_1 - u_1v_3, \ u_1v_2 - u_2v_1).$$
Here's a way to remember it: the first entry leaves out the first entries of $u$ and $v$ and crosses the other two ($u_2v_3 - u_3v_2$), and each next entry shifts every index up by one, wrapping $3$ around to $1$.
Find $(1, 2, 3) \times (4, 5, 6)$, and check that it's orthogonal to both vectors.
$(2 \cdot 6 - 3 \cdot 5, \ 3 \cdot 4 - 1 \cdot 6, \ 1 \cdot 5 - 2 \cdot 4) = (12 - 15, \ 12 - 6, \ 5 - 8) = (-3, 6, -3)$.
Check: $(-3, 6, -3) \cdot (1, 2, 3) = -3 + 12 - 9 = 0$ and $(-3, 6, -3) \cdot (4, 5, 6) = -12 + 30 - 18 = 0$.
Here's the answer standing off the sheet the two vectors lie on. The camera circles so you can see both right angles.
That check worked because it always works. Here's why the formula is shaped the way it is.
For all $u, v$ in $\R^3$, $u \times v$ is orthogonal to both $u$ and $v$.
$u \cdot (u \times v) = u_1(u_2v_3 - u_3v_2) + u_2(u_3v_1 - u_1v_3) + u_3(u_1v_2 - u_2v_1)$. Multiply out: the six terms are $u_1u_2v_3 - u_1u_3v_2 + u_2u_3v_1 - u_1u_2v_3 + u_1u_3v_2 - u_2u_3v_1$, and they cancel in pairs, leaving $0$. The same computation with $v$ in place of the first $u$ gives $v \cdot (u \times v) = 0$. $\blacksquare$
This one is the computation itself, drawn as cards. Multiply out $u \cdot (u \times v)$ and you get six terms. Each one has a partner made of the same three letters with the opposite sign:
Every pair cancels, so nothing is left: $u \cdot (u \times v) = 0$. The same six cards with $v$ in place of the first $u$ cancel the same way, so $v \cdot (u \times v) = 0$ too. That's the secret behind the random-looking formula: it's built so the cards come in pairs. The written-out version is in the n version.
A product that hands back a vector perpendicular to both inputs only works in $\R^3$... and, weirdly, in $\R^7$. Nobody ever invites $\R^7$ to anything else.
Orthogonal to both still leaves two choices: up or down. And how long is it? Watch $v$ swing around $u$ on the floor:
The cross product stands straight up off the floor, as long as the shaded parallelogram's area. As $v$ passes $180^\circ$, the area flattens to $0$ and the cross product flips to point down. The direction follows the right-hand rule: curl the fingers of your right hand from $u$ toward $v$, and your thumb points along $u \times v$. Now the length, exactly.
If $\theta$ is the angle between the nonzero vectors $u$ and $v$ in $\R^3$, then $$\|u \times v\| = \|u\| \, \|v\| \sin\theta,$$ the area of the parallelogram with sides $u$ and $v$. In particular, $u \times v = 0$ exactly when $u$ and $v$ are parallel (or one of them is $0$).
Square every entry of $u \times v$ and add. Expanding and regrouping (it's a page of algebra; try it once) gives Lagrange's identity: $$\|u \times v\|^2 = (u_1^2 + u_2^2 + u_3^2)(v_1^2 + v_2^2 + v_3^2) - (u_1v_1 + u_2v_2 + u_3v_3)^2 = \|u\|^2\|v\|^2 - (u \cdot v)^2.$$ By the angle formula, $(u \cdot v)^2 = \|u\|^2\|v\|^2\cos^2\theta$, so $\|u \times v\|^2 = \|u\|^2\|v\|^2(1 - \cos^2\theta) = \|u\|^2\|v\|^2\sin^2\theta$. Since $0^\circ \le \theta \le 180^\circ$, $\sin\theta \ge 0$, and taking square roots gives the formula. A parallelogram with sides $\|u\|$ and $\|v\|$ meeting at angle $\theta$ has base $\|u\|$ and height $\|v\|\sin\theta$, so that's its area. Finally, $u \times v = 0$ exactly when $\sin\theta = 0$, that is, $\theta = 0^\circ$ or $180^\circ$: parallel. $\blacksquare$
This is a visual argument for the area and a numerical check for the rest. First the area: drop a height from $v$'s tip to $u$'s line. The right triangle it makes has hypotenuse $\|v\|$ and angle $\theta$, so the height is $\|v\|\sin\theta$ (SOH-CAH-TOA). Now cut that triangle off one end and slide it to the other:
The parallelogram becomes a rectangle with base $\|u\|$ and height $\|v\|\sin\theta$, so its area is $\|u\| \, \|v\|\sin\theta$. That's the right side of the formula, honestly pictured. That the cross product's length matches it is a page of algebra called Lagrange's identity, and no picture here shows that part. The animation's last step checks it on the example above: $\|(-3, 6, -3)\| = \sqrt{54}$, and $\|u\| \, \|v\|\sin\theta$ for $(1, 2, 3)$ and $(4, 5, 6)$ comes out to $\sqrt{54}$ too. And a flat parallelogram (parallel sides, $\theta = 0^\circ$ or $180^\circ$) has no area, so its cross product is $0$. The algebra is in the n version.
Before you trust the right-hand rule, test it. Each round shows two vectors and two arrows standing off their plane; one of them is $u \times v$:
The order matters, and that's the first of the rules on the cross product's list. Most of them look like ordinary multiplication. One very much doesn't.
For all $u, v, w$ in $\R^3$ and every real number $a$:
All five come straight from the formula. (1) Swapping $u$ and $v$ swaps the two products in each entry, which flips each entry's sign. (2) Then $u \times u = -(u \times u)$, so it's $0$. (3) Every entry is a difference of products with one factor from $u$ and one from $v$, so a factor of $a$ on either one comes out front. (4) Each entry is linear in each vector's entries, so it distributes. (5) For example, $e_1 \times e_2 = (0 \cdot 0 - 0 \cdot 1, \ 0 \cdot 0 - 1 \cdot 0, \ 1 \cdot 1 - 0 \cdot 0) = (0, 0, 1) = e_3$; the other two are the same computation. $\blacksquare$
This is a visual argument for four of the five, not a formal proof. Watch the right-hand rule run around the corner where the axes meet:
(5) Curl your fingers from $e_1$ to $e_2$ and your thumb points up $e_3$; move every name one step around the cycle $1 \to 2 \to 3 \to 1$ and the same hand gives the other two. (1) Curl the other way, from $v$ to $u$, and your thumb flips over: $v \times u$ has the same length (same parallelogram) and the opposite direction. (2) $u$ with itself makes a parallelogram with no area, and an arrow of length $0$ is $0$. (3) Stretch one side by $a$ and the area stretches by $|a|$; a negative $a$ also turns that side around, and the thumb flips with it. (4) is the one with no quick picture: it holds because every entry of the formula is a difference of products with one factor from each vector, and those distribute like ordinary multiplication. The full proof is in the n version.
What's missing from that list is associativity, and it's missing because it's false. $(e_1 \times e_1) \times e_2 = 0 \times e_2 = 0$, but $e_1 \times (e_1 \times e_2) = e_1 \times e_3 = -e_2$. Where the parentheses go changes the answer.
Every multiplication I've met so far ignored the parentheses. This one reads them. Personally.
The middle entry is the trap: it's $u_3v_1 - u_1v_3$, and almost everyone writes $u_1v_3 - u_3v_1$ the first time, which flips its sign. The orthogonality check catches it every time, so do the check. And $u \times v$ versus $v \times u$ is a sign flip too, so keep the order the question gives you.
Your turn. Check your answer against both vectors before you pick.
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The length theorem turns areas into a cross product and one square root. Drag the slider and watch the two sides of the equation agree:
For a triangle, take half: the triangle with corners $A$, $B$, $C$ is half the parallelogram on the sides $B - A$ and $C - A$.
Find the area of the triangle with corners $A = (1, 0, 0)$, $B = (0, 2, 0)$, and $C = (0, 0, 3)$.
The sides from $A$ are $B - A = (-1, 2, 0)$ and $C - A = (-1, 0, 3)$. Their cross product is $(2 \cdot 3 - 0 \cdot 0, \ 0 \cdot (-1) - (-1) \cdot 3, \ (-1) \cdot 0 - 2 \cdot (-1)) = (6, 3, 2)$, with length $\sqrt{36 + 9 + 4} = 7$. That's the parallelogram, so the triangle's area is $\frac{7}{2}$.
Here it is in space: the triangle on the three axes, doubled into a parallelogram, with the cross product standing on it.
Your turn: parallelograms from two sides, triangles from three corners.
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One more step up. Three vectors $u$, $v$, $w$ from one corner build a slanted box, a parallelepiped. Its base is the parallelogram on $v$ and $w$, and its height is how far $u$ reaches off that base:
The base's area is $\|v \times w\|$, and the height is the length of $u$'s shadow on the line through $v \times w$, which is $\frac{|u \cdot (v \times w)|}{\|v \times w\|}$. Multiply, and the $\|v \times w\|$ cancels.
The parallelepiped with edges $u$, $v$, $w$ in $\R^3$ has volume $|u \cdot (v \times w)|$. In particular, $u \cdot (v \times w) = 0$ exactly when $u$, $v$, $w$ lie in one plane through $0$.
If $v$ and $w$ are parallel, the box is flat, $v \times w = 0$, and both sides are $0$. Otherwise the base has area $\|v \times w\|$ and $n = v \times w$ is orthogonal to the base. The height is the length of the projection of $u$ onto $n$: $\left\|\frac{u \cdot n}{n \cdot n} n\right\| = \frac{|u \cdot n|}{\|n\|^2}\|n\| = \frac{|u \cdot n|}{\|n\|}$. Base times height is $|u \cdot n|$. The volume is $0$ exactly when the height is $0$ (with $v$, $w$ not parallel), that is, when $u$ lies in the plane spanned by $v$ and $w$. $\blacksquare$
This is a visual argument, not a formal proof. It leans on a fact from geometry class, Cavalieri's principle: sliding the layers of a solid sideways, like pushing a deck of cards askew, doesn't change its volume. So slide the slanted box's top until it sits straight above the base:
Now it's a straight box: base times height. The base is the parallelogram on $v$ and $w$, with area $\|v \times w\|$. The height is how far $u$ reaches along $v \times w$, the arrow standing straight up off the base, which is the length of $u$'s shadow on that arrow, $\frac{|u \cdot (v \times w)|}{\|v \times w\|}$. Multiply, and the $\|v \times w\|$ cancels. And when $u$ tips down into the base's plane, the height is $0$, and so is the volume. The full proof is in the n version.
"Lie in one plane through $0$" is the picture of linear dependence for three vectors in $\R^3$. So a single number, $u \cdot (v \times w)$, tests whether three vectors are linearly independent: nonzero means yes.
So $u \cdot (v \times w)$ is called the triple product, and it's the dot product and the cross product working together.
(a) Find the volume of the box with edges $u = (1, 0, 2)$, $v = (0, 3, 1)$, $w = (2, 1, 0)$. (b) For which $k$ do $(1, 2, k)$, $(0, 3, 1)$, and $(2, 1, 0)$ lie in one plane through $0$?
(a) $v \times w = (3 \cdot 0 - 1 \cdot 1, \ 1 \cdot 2 - 0 \cdot 0, \ 0 \cdot 1 - 3 \cdot 2) = (-1, 2, -6)$, so $u \cdot (v \times w) = -1 + 0 - 12 = -13$. The volume is $|-13| = 13$.
(b) With the same $v \times w$, $(1, 2, k) \cdot (-1, 2, -6) = -1 + 4 - 6k = 3 - 6k$. That's $0$ when $k = \frac{1}{2}$.
Both parts, moving: first the box from (a), then $u = (1, 2, k)$ as $k$ slides. At $k = \frac{1}{2}$ the box goes flat, because the third edge lies in the plane of the other two.
The triple product is the $3 \times 3$ determinant with rows $u$, $v$, $w$, and the "volume of the box" is exactly what determinants measure in general.
Your turn. Volumes are never negative, so take the absolute value at the end.
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That's the toolkit: the dot product for lengths, angles, and shadows, the cross product for perpendicular directions and areas, and the two together for volume. The next page, Lines & Planes in Space, puts every one of them to work.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Figuring out which product a question needs is half the work.
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Made it this far? Time for something with a little heat. Find a vector with a given length that's orthogonal to two others and points to a named side. It takes a cross product, a length, a rescale, and a sign check.
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