Back to the on-screen lesson ·
A dot and a cross in one expression: the determinant of three vectors, its absolute value as the volume of a parallelepiped, its sign as handedness, and its vanishing as the test for coplanarity.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute a scalar triple product as a three by three determinant, read its absolute value as the volume of the box the three vectors span and its sign as their handedness, and use its vanishing to test three vectors for coplanarity or four points for lying on one plane. You will also be able to say which reorderings of the three vectors are free and which change the sign.
The cross product, whose length is the area of a parallelogram, and the dot product, which keeps the part of one vector along another. This lesson puts them together in one expression, and the combination turns out to compute a volume — with a sign attached that carries real information.
A parallelepiped is the slanted box three vectors span from a common corner. Its signed volume is the triple product, positive when the three vectors are right-handed in the order given and negative when they are left-handed. Three vectors are coplanar when one plane through their common tail contains all three. A cyclic permutation rotates the three vectors without reversing their order: $\mathbf{u}, \mathbf{v}, \mathbf{w}$ becomes $\mathbf{v}, \mathbf{w}, \mathbf{u}$.
The scalar triple product of three vectors is
$$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}.$$
The determinant is not a mnemonic this time — unlike the cross product's, every entry here is a number, so the array really is a determinant and every fact about determinants applies.
What it measures. $\mathbf{v} \times \mathbf{w}$ is perpendicular to the base parallelogram and its length is that parallelogram's area. Dotting with $\mathbf{u}$ keeps the part of $\mathbf{u}$ along that perpendicular — which is the height of the box above the base. So
$$|\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| = \text{area of base} \times \text{height} = \text{volume}.$$
The sign. Without the absolute value the number is a signed volume: positive when $\mathbf{u}$ lies on the same side of the base as $\mathbf{v} \times \mathbf{w}$, which is what it means for the three to be right-handed in that order.
Symmetry. Rotating the three vectors cyclically leaves the value alone:
$$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) = \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}),$$
and swapping any two of them negates it. Both facts are the determinant's row rules in disguise. A useful consequence: the dot and the cross may be exchanged, $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}$, so the brackets are the only thing that has to be right.
The test. The triple product is zero exactly when the three vectors are coplanar — a flat box has no volume. That is the three-vector analogue of the cross product's parallel test, and it is how this course will decide whether four points lie on one plane and whether three directions are independent.
Another way: picture
Stand a slanted box on a table. Its base is the parallelogram $\mathbf{v}$ and $\mathbf{w}$ span, and $\mathbf{u}$ is the leaning edge from the same corner. Volume is base area times vertical height, and the lean contributes nothing — push the top of the box sideways without lifting it and the volume does not change at all. That is why adding a multiple of one edge to another leaves the determinant alone.
Another way: steps
To decide whether four points lie on one plane:
Every property of a $3 \times 3$ determinant reads as a statement about the box.
Swapping two rows negates it. Exchanging two edges reflects the box, which cannot change how big it is and does reverse its handedness. So the magnitude survives and the sign flips.
A repeated row makes it zero. Two identical edges leave the box flat, with no volume.
Adding a multiple of one row to another changes nothing. That is the shear in the picture above: sliding the top face sideways over the base.
Scaling a row scales the determinant. Stretching one edge by $c$ stretches the box by $c$.
This is worth more than it looks. It means the determinant is not an arbitrary formula that happens to detect degeneracy; it is the volume, and every algebraic manipulation you are allowed to perform on it is a motion of the box that a bricklayer would call obvious. Linear algebra builds the general theory on exactly this footing.
Taking the absolute value too early. The signed value is what carries the handedness; throw it away before you have read it and you have thrown away half the answer.
Moving the brackets. $(\mathbf{u} \cdot \mathbf{v}) \times \mathbf{w}$ is meaningless: the left factor is a number, and a number cannot be crossed with anything. Only $\cdot$ and $\times$ may be exchanged, never the bracketing.
Reordering the rows at will. A cyclic rotation is free; a single swap negates. Two different orderings of the same three vectors give answers that differ by a sign, and in an orientation problem that sign is the answer.
Expecting exact zero from measured data. Real vectors give a small non-zero triple product, and the useful question is how small compared with the product of the three lengths. An orientation, an order of integration and an order of factors are part of the answer, not part of the handwriting. Reversing a curve flips the sign of the work along it, swapping the factors of a cross product flips the vector, and exchanging the two limits of an inner integral changes what region was integrated over. Say which one you chose, every time.
$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is a scalar. There is also a vector triple product, $\mathbf{u} \times (\mathbf{v} \times \mathbf{w})$, which is a completely different object: it is a vector, it lies in the plane of $\mathbf{v}$ and $\mathbf{w}$, and it is not associative, so $(\mathbf{u} \times \mathbf{v}) \times \mathbf{w}$ is usually something else again.
So the brackets and the symbols together decide what you have. Read the expression before computing: a dot anywhere in it means the answer is a number, and a misplaced bracket in a page of vector algebra is not a typing slip but a different quantity that happens to be spelled almost the same.
Take $\mathbf{u} = \langle 1, 0, 0 \rangle$, $\mathbf{v} = \langle 1, 2, 0 \rangle$ and $\mathbf{w} = \langle 1, 2, 3 \rangle$ from one corner.
Three edges, written as rows.
The array is triangular, so the determinant is the product of the diagonal: $1 \times 2 \times 3 = 6$.
Every entry above the diagonal is zero.
The volume is $6$, and the sign is positive, so the three edges are right-handed in this order. The off-diagonal $1$s described how far the box leans and contributed nothing — the shear rule, seen in arithmetic.
Leaning a box does not change its volume.
Are $(0,0,0)$, $(1,2,3)$, $(2,3,4)$ and $(3,5,7)$ on one plane? Subtract the first from the others: $\langle 1,2,3 \rangle$, $\langle 2,3,4 \rangle$, $\langle 3,5,7 \rangle$.
Three edges from a common corner.
The third is the sum of the first two, so one row of the array is the sum of the other two and the determinant is zero.
A dependent row makes the box flat.
So the four points are coplanar. Spotting the dependency was quicker than expanding, and it also said why: the fourth point was reachable from the first three without ever leaving their plane.
Zero volume is the coplanarity test.
The three edges from the origin are $\langle 2,0,0 \rangle$, $\langle 0,3,0 \rangle$ and $\langle 0,0,4 \rangle$.
The corner at the origin makes the edges the points themselves.
The array is diagonal, so the triple product is $2 \times 3 \times 4 = 24$ and the parallelepiped has volume $24$.
The tetrahedron is one sixth of that, so its volume is $4$. Worth checking against the elementary formula: base area $\tfrac12 \times 2 \times 3 = 3$ and height $4$ give $\tfrac13 \times 3 \times 4 = 4$ as well, which is where the sixth comes from — a half and a third.
Write the $3 \times 3$ array whose determinant is $\mathbf{v} \cdot (\mathbf{w} \times \mathbf{u})$, for $\mathbf{u} = \langle 4, -1, -1 \rangle$, $\mathbf{v} = \langle -3, 0, -4 \rangle$ and $\mathbf{w} = \langle 4, -3, -4 \rangle$.
This task has no paper form; do it on a device.
For which values of $k$ are $\langle 1, 0, 0 \rangle$, $\langle 0, 1, 2 \rangle$ and $\langle 3, 4, k \rangle$ coplanar? Give the set of $k$.
This task has no paper form; do it on a device.
Find the volume of the parallelepiped with edges $\langle 2, 0, 0 \rangle$, $\langle -3, 5, 0 \rangle$ and $\langle 2, 2, 3 \rangle$.
Answer:
Find the volume of the tetrahedron with edges $\langle 6, 0, 0 \rangle$, $\langle -3, 9, 0 \rangle$ and $\langle 3, 2, 3 \rangle$ from one corner.
Answer:
Write the $3 \times 3$ array whose determinant is $\mathbf{v} \cdot (\mathbf{w} \times \mathbf{u})$, for $\mathbf{u} = \langle -1, 2, -4 \rangle$, $\mathbf{v} = \langle -2, 0, -1 \rangle$ and $\mathbf{w} = \langle 4, 1, 4 \rangle$.
This task has no paper form; do it on a device.
A sensor rig reports three non-zero displacement vectors, no two of them parallel, whose scalar triple product is $0$ to within $4$ decimal places. What follows?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For which values of $k$ are $\langle 1, 0, 3 \rangle$, $\langle 0, 1, -1 \rangle$ and $\langle -3, 2, k \rangle$ coplanar? Give the set of $k$.
This task has no paper form; do it on a device.
You can compute a triple product, read it as a signed volume, and use it to test three vectors for lying in one plane. Next: the simplest object in space that a point and a direction describe, which is a line.
10. Your turn: the volume of the tetrahedron with corners at the origin, $(2,0,0)$, $(0,3,0)$ and $(0,0,4)$, step 3