Precalculus

The cross product

Two arrows in space span a flat parallelogram. Is there a single calculation that returns an arrow standing straight up from it, as long as the parallelogram’s area?

You need

Orthogonal vectorsPrecalculus Vectors in spacePrecalculus Computing a determinantPrecalculus Area from two sides and the included anglePrecalculus +1 more

Where you are 10 in Polar coordinates and vectors

Trigonometry7

Polar coordinates and vectors10

▴ 5 above Eliminating the parameter Polar equations and their graphs3 The dot product3 Polar form of a complex number3

The cross product

Why this exists

The dot product turns two vectors into a number that measures how much they point the same way. Many questions in space ask for something else. Which direction is straight out of a tilted roof panel? How large is a triangle whose corners are three points in space? Is a third point on the line through two others?

One operation, which exists only in three dimensions, answers all three. The cross product takes vectors \mathbf{u} and \mathbf{v} in space and returns a vector \mathbf{u} \times \mathbf{v} perpendicular to both, whose length is the area of the parallelogram they span and whose direction is fixed by the right-hand rule of Vectors in space.

Definition

For \mathbf{u} = \langle u_1, u_2, u_3 \rangle and \mathbf{v} = \langle v_1, v_2, v_3 \rangle, the cross product is the vector

\begin{aligned} \mathbf{u} \times \mathbf{v} = \langle\, &u_2 v_3 - u_3 v_2, \\ &u_3 v_1 - u_1 v_3, \\ &u_1 v_2 - u_2 v_1 \,\rangle. \end{aligned}

The pattern is easiest to carry as a 3 \times 3 determinant with \mathbf{i}, \mathbf{j}, \mathbf{k} in the first row, \mathbf{u} in the second and \mathbf{v} in the third:

\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}

Expanding along the first row, with the checkerboard signs +, -, + of the cofactor expansion, gives three 2 \times 2 determinants:

\begin{aligned} \mathbf{u} \times \mathbf{v} ={}& \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix}\mathbf{i} \\ &- \begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix}\mathbf{j} \\ &+ \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix}\mathbf{k}. \end{aligned}

Each block is what remains after crossing out the column of that unit vector, and the minus sign on the middle term gives the middle component above. The first row holds vectors, not numbers, so this “determinant” is a memory device, not a true determinant.

The dot product gives a number; the cross product gives a vector. It is defined only for vectors with three components.

Intuition

Figure 1The height over the base \mathbf{u} is \|\mathbf{v}\|\sin\theta, so the area is \|\mathbf{u}\|\,\|\mathbf{v}\|\sin\theta.

Two vectors from one point, \mathbf{u} and \mathbf{v} with angle \theta between them, are two sides of a parallelogram. Take \mathbf{u} as the base. The height is the leg opposite \theta in a right triangle with hypotenuse \|\mathbf{v}\|, so it is \|\mathbf{v}\|\sin\theta, and the area is

\text{area} = \|\mathbf{u}\|\,\|\mathbf{v}\|\sin\theta.

The triangle with sides \mathbf{u} and \mathbf{v} is half of it, which is the formula of Area from two sides and the included angle. The cross product packages this area together with a direction: the one straight out of the parallelogram.

Figure 2For \mathbf{u} = \langle 0, 3, 0 \rangle and \mathbf{v} = \langle -2, 1, 0 \rangle in the floor, \mathbf{u} \times \mathbf{v} = \langle 0, 0, 6 \rangle. It stands straight up, and its length 6 is the area of the shaded parallelogram.

In the second figure the parallelogram lies in the floor, so “straight out” means up or down. The first two components are 0, since each product in them has a factor u_3 = 0 or v_3 = 0, and the third is 0 \cdot 1 - 3 \cdot (-2) = 6: straight up, with length 6, the base 3 times the height 2.

Results

For vectors \mathbf{u}, \mathbf{v}, \mathbf{w} in space and any number c:

Perpendicular to both. (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{u} = 0 and (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{v} = 0, so a nonzero \mathbf{u} \times \mathbf{v} is orthogonal to both \mathbf{u} and \mathbf{v}.

Length is area. For nonzero \mathbf{u} and \mathbf{v} with angle \theta between them, 0^\circ \le \theta \le 180^\circ,

\|\mathbf{u} \times \mathbf{v}\| = \|\mathbf{u}\|\,\|\mathbf{v}\|\sin\theta,

the area of the parallelogram with sides \mathbf{u} and \mathbf{v}. The triangle with those sides has half that area.

Direction: the right-hand rule. Point the fingers of your right hand along \mathbf{u} and curl them toward \mathbf{v} through the angle \theta. Your thumb points along \mathbf{u} \times \mathbf{v}. In particular \mathbf{i} \times \mathbf{j} = \mathbf{k}, \mathbf{j} \times \mathbf{k} = \mathbf{i} and \mathbf{k} \times \mathbf{i} = \mathbf{j}: around the cycle \mathbf{i} \to \mathbf{j} \to \mathbf{k} \to \mathbf{i}, forward gives the next letter and backward gives its negative.

Order matters. \mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v}).

Parallel vectors. \mathbf{u} \times \mathbf{u} = \mathbf{0}. For nonzero \mathbf{u} and \mathbf{v}, \mathbf{u} \times \mathbf{v} = \mathbf{0} if and only if \mathbf{u} and \mathbf{v} are parallel (one is a multiple of the other).

Scaling and adding. (c\mathbf{u}) \times \mathbf{v} = c(\mathbf{u} \times \mathbf{v}) = \mathbf{u} \times (c\mathbf{v}), and (\mathbf{u} + \mathbf{w}) \times \mathbf{v} = \mathbf{u} \times \mathbf{v} + \mathbf{w} \times \mathbf{v}, with the same rule in the second slot.

Why it is true

Perpendicular. Dot \mathbf{u} with the component formula:

\begin{aligned} &u_1(u_2 v_3 - u_3 v_2) + u_2(u_3 v_1 - u_1 v_3) \\ &\quad + u_3(u_1 v_2 - u_2 v_1). \end{aligned}

Multiplied out (the The distributive property), the six terms cancel in pairs: u_1u_2v_3 with -u_1u_2v_3, and likewise the other two pairs. The same cancellation happens with \mathbf{v}.

Length. The key step is an identity that holds for every pair of vectors:

\begin{aligned} \|\mathbf{u} \times \mathbf{v}\|^2 &= \|\mathbf{u}\|^2\|\mathbf{v}\|^2 \\ &\quad - (\mathbf{u} \cdot \mathbf{v})^2. \end{aligned}

On the right, \|\mathbf{u}\|^2\|\mathbf{v}\|^2 has the nine terms u_i^2 v_j^2, and (u_1v_1 + u_2v_2 + u_3v_3)^2 has the three squares u_i^2 v_i^2 plus 2u_iv_iu_jv_j for each pair i < j. The squares cancel, and for each pair i < j what remains is u_i^2v_j^2 + u_j^2v_i^2 - 2u_iu_jv_iv_j = (u_iv_j - u_jv_i)^2, the square of one component of \mathbf{u} \times \mathbf{v}. The three pairs give the left side.

Now use \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta from The dot product and the angle between vectors:

\begin{aligned} \|\mathbf{u} \times \mathbf{v}\|^2 &= \|\mathbf{u}\|^2\|\mathbf{v}\|^2(1 - \cos^2\theta) \\ &= \|\mathbf{u}\|^2\|\mathbf{v}\|^2\sin^2\theta. \end{aligned}

On [0^\circ, 180^\circ] the sine is never negative, so the square root is \|\mathbf{u}\|\,\|\mathbf{v}\|\sin\theta, the area found in Intuition.

Order and parallel vectors. Swapping \mathbf{u} and \mathbf{v} swaps the two rows of each 2 \times 2 block, and cb - da = -(ad - bc), so every component changes sign. With \mathbf{v} = \mathbf{u}, each block has two equal rows, so it is 0. For nonzero vectors the length formula gives \mathbf{0} exactly when \sin\theta = 0, that is \theta = 0^\circ or 180^\circ: parallel.

Scaling and adding. Each component has the form ad - bc with (a, b) taken from \mathbf{u}. Replacing \mathbf{u} by c\mathbf{u} multiplies ad - bc by c. Replacing \mathbf{u} by \mathbf{u} + \mathbf{w} splits each ad - bc into the sum of the two matching expressions.

Direction. The formula gives \mathbf{i} \times \mathbf{j} = \mathbf{k} (components 0, 0, 1), matching the right-hand system the axes were built on, and likewise \mathbf{j} \times \mathbf{k} = \mathbf{i}, \mathbf{k} \times \mathbf{i} = \mathbf{j}. As \mathbf{u} and \mathbf{v} turn, the components change continuously, so the product could flip to the opposite side only by passing through \mathbf{0}, which needs parallel vectors.

Worked examples

Example 1

Computing and checking

Find \mathbf{u} \times \mathbf{v} for \mathbf{u} = \langle 2, -1, 3 \rangle and \mathbf{v} = \langle 1, 4, -2 \rangle, and check that the answer is perpendicular to both.

Solution

Set up the determinant and expand along the first row:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 4 & -2 \end{vmatrix}

The \mathbf{i} block is (-1)(-2) - (3)(4) = -10. The \mathbf{j} block is (2)(-2) - (3)(1) = -7, and its sign is minus, giving +7. The \mathbf{k} block is (2)(4) - (-1)(1) = 9. So \mathbf{u} \times \mathbf{v} = \langle -10, 7, 9 \rangle.

Check with dot products: with \mathbf{u}, -20 - 7 + 27 = 0; with \mathbf{v}, -10 + 28 - 18 = 0. A sign slip in any block would almost always break one of these.

Example 2

Area two ways

Find the area of the parallelogram spanned by \mathbf{u} = \langle 1, 4, 8 \rangle and \mathbf{v} = \langle 4, 0, 3 \rangle. Then find the angle between them from the dot product and confirm that \|\mathbf{u}\|\,\|\mathbf{v}\|\sin\theta gives the same area.

Solution

The components are 4 \cdot 3 - 8 \cdot 0 = 12, 8 \cdot 4 - 1 \cdot 3 = 29 and 1 \cdot 0 - 4 \cdot 4 = -16, so \mathbf{u} \times \mathbf{v} = \langle 12, 29, -16 \rangle. The area is its length:

\sqrt{144 + 841 + 256} = \sqrt{1241} \approx 35.2.

Second route: \displaystyle \|\mathbf{u}\| = \sqrt{81} = 9, \displaystyle \|\mathbf{v}\| = \sqrt{25} = 5 and \mathbf{u} \cdot \mathbf{v} = 4 + 0 + 24 = 28, so \displaystyle \cos\theta = \frac{28}{45} and \theta \approx 51.5^\circ. Then \displaystyle \sin^2\theta = 1 - \frac{784}{2025} = \frac{1241}{2025}, so \displaystyle 45\sin\theta = \sqrt{1241}, the same area.

Example 3

A tarp on three poles

Hjalmar stretches a triangular tarp between the tops of three poles at a campsite. In meters, with z measured up, the corners are A(0, 0, 3), B(6, 0, 4) and C(2, 5, 3). Find the area of the tarp and a vector perpendicular to it that points upward.

Solution

Two sides from the shared corner A: \overrightarrow{AB} = \langle 6, 0, 1 \rangle and \overrightarrow{AC} = \langle 2, 5, 0 \rangle. Their cross product has components 0 \cdot 0 - 1 \cdot 5 = -5, 1 \cdot 2 - 6 \cdot 0 = 2 and 6 \cdot 5 - 0 \cdot 2 = 30:

\overrightarrow{AB} \times \overrightarrow{AC} = \langle -5, 2, 30 \rangle.

The tarp is half of the parallelogram on these two sides, so its area is

\tfrac{1}{2}\sqrt{25 + 4 + 900} = \tfrac{1}{2}\sqrt{929} \approx 15.24

square meters. The vector \langle -5, 2, 30 \rangle is perpendicular to both sides (check: -30 + 30 = 0 and -10 + 10 = 0), so to the whole tarp, and its positive third component means it points upward.

Example 4

Working from a known cross product

Baldur knows only that \mathbf{u} \times \mathbf{v} = \langle 2, -6, 3 \rangle. (a) Find \mathbf{v} \times \mathbf{u}, (2\mathbf{u}) \times \mathbf{v} and (\mathbf{u} + \mathbf{v}) \times \mathbf{v}. (b) Find the area of the triangle with sides \mathbf{u} and \mathbf{v}. (c) He also learns \|\mathbf{u}\| = 2 and \|\mathbf{v}\| = 5. What can the angle between them be?

Solution

(a) Reversing the order negates: \mathbf{v} \times \mathbf{u} = \langle -2, 6, -3 \rangle. Scaling one factor scales the product: (2\mathbf{u}) \times \mathbf{v} = \langle 4, -12, 6 \rangle. Adding splits the product, and \mathbf{v} \times \mathbf{v} = \mathbf{0}:

\begin{aligned} (\mathbf{u} + \mathbf{v}) \times \mathbf{v} &= \mathbf{u} \times \mathbf{v} + \mathbf{0} \\ &= \langle 2, -6, 3 \rangle. \end{aligned}

(b) \displaystyle \|\mathbf{u} \times \mathbf{v}\| = \sqrt{4 + 36 + 9} = 7, so the triangle has area \displaystyle \frac{7}{2}.

(c) 7 = 2 \cdot 5 \sin\theta gives \sin\theta = 0.7. Two angles in [0^\circ, 180^\circ] have that sine: \theta \approx 44.4^\circ or \theta \approx 135.6^\circ. The cross product alone cannot choose; the sign of \mathbf{u} \cdot \mathbf{v} would.

Example 5

A unit vector perpendicular to two others

Eyvindur needs a unit vector perpendicular to both \mathbf{a} = \langle 1, 0, -1 \rangle and \mathbf{b} = \langle 0, 2, -1 \rangle.

Solution

The components of \mathbf{a} \times \mathbf{b} are 0 \cdot (-1) - (-1)(2) = 2, (-1)(0) - (1)(-1) = 1 and 1 \cdot 2 - 0 \cdot 0 = 2, so \mathbf{a} \times \mathbf{b} = \langle 2, 1, 2 \rangle, of length \displaystyle \sqrt{9} = 3. Dividing by 3:

\left\langle \tfrac{2}{3}, \tfrac{1}{3}, \tfrac{2}{3} \right\rangle \text{ or its negative.}

Both work: a line has two unit directions. Check: \displaystyle \frac{2}{3} - \frac{2}{3} = 0 for each of \mathbf{a} and \mathbf{b}.

Example 6

Shadows of a slanted triangle

The triangle with corners P(2, 0, 0), Q(0, 3, 0) and R(0, 0, 6) cuts across the corner of the first octant. Find its area, and compare its square with the squares of the areas of its three shadows on the coordinate planes.

Solution

From P: \overrightarrow{PQ} = \langle -2, 3, 0 \rangle and \overrightarrow{PR} = \langle -2, 0, 6 \rangle. The components are 3 \cdot 6 - 0 \cdot 0 = 18, 0 \cdot (-2) - (-2)(6) = 12 and (-2)(0) - (3)(-2) = 6, so the cross product is \langle 18, 12, 6 \rangle = 6\langle 3, 2, 1 \rangle, of length \displaystyle 6\sqrt{14}. The area is half: \displaystyle 3\sqrt{14} \approx 11.22.

Dropping the triangle straight onto each coordinate plane gives a right triangle with legs on the axes: in the xy-plane legs 2 and 3 (area 3), in the xz-plane legs 2 and 6 (area 6), in the yz-plane legs 3 and 6 (area 9). Then

3^2 + 6^2 + 9^2 = 9 + 36 + 81 = 126,

and \displaystyle (3\sqrt{14})^2 = 126 as well: the shadow areas 9, 6, 3 are half the components 18, 12, 6, and the length formula adds their squares. It is a Pythagorean theorem for areas.

Boundaries

Three dimensions only. The formula needs exactly three components. Two plane vectors can be crossed only after placing them in the floor as \langle a, b, 0 \rangle and \langle c, d, 0 \rangle; the result is \langle 0, 0, ad - bc \rangle, the 2 \times 2 determinant.
Not commutative, not associative. \mathbf{v} \times \mathbf{u} is the negative of \mathbf{u} \times \mathbf{v}, and (\mathbf{i} \times \mathbf{i}) \times \mathbf{j} = \mathbf{0} while \mathbf{i} \times (\mathbf{i} \times \mathbf{j}) = -\mathbf{j}.
No canceling. \mathbf{u} \times \mathbf{v} = \mathbf{u} \times \mathbf{w} does not force \mathbf{v} = \mathbf{w}: \mathbf{i} \times \mathbf{i} = \mathbf{i} \times (2\mathbf{i}) = \mathbf{0}.
A zero product says only “parallel,” not whether the vectors point the same way or opposite ways.
The length gives the sine, not the angle. \theta and 180^\circ - \theta have the same sine; the dot product’s sign tells acute from obtuse.
The direction depends on handedness. In a left-hand system the same formula points the other way.

Where this goes next

A vector perpendicular to a flat surface is what is needed to write the equation of a plane through three points, the subject of Lines and planes in space. The number \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) is the 3 \times 3 determinant of Determinants by cofactor expansion, and its absolute value is the volume of the slanted box on the three vectors. In physics it measures the turning effect (torque) of a force.

Exercises

Computing cross products

Find \langle 1, 3, -2 \rangle \times \langle 4, 0, 1 \rangle.
Find \langle 0, 2, 5 \rangle \times \langle 3, -1, 0 \rangle and check with dot products that it is perpendicular to both.
Use the products of the unit vectors and the adding rule to find (\mathbf{i} + \mathbf{j}) \times \mathbf{k}.
Find \langle 3, -1, 2 \rangle \times \langle -6, 2, -4 \rangle and explain the result.

Length and area

Find the area of the parallelogram spanned by \langle 2, 2, -1 \rangle and \langle 1, -2, 3 \rangle.
Find the area of the triangle with vertices P(1, 1, 0), Q(3, 1, 2) and R(1, 4, 1).
\|\mathbf{u}\| = 4, \|\mathbf{v}\| = 6 and the angle between them is 150^\circ. Find \|\mathbf{u} \times \mathbf{v}\| and \mathbf{u} \cdot \mathbf{v}.
Use the cross product to find \sin\theta for \langle 1, 0, 1 \rangle and \langle 0, 1, 1 \rangle, then use the dot product to decide \theta.
Hjalmar’s second tarp is a parallelogram ABCD with A(0, 0, 1), B(4, 0, 1) and D(1, 3, 5), in meters. Find C and the area of the tarp.
Find the area of the flat triangle with vertices (0, 0), (6, 1) and (2, 5) by placing it in the floor z = 0.

Direction and algebra

Without the formula, find \mathbf{j} \times \mathbf{i}, \mathbf{k} \times (-\mathbf{j}) and (2\mathbf{i}) \times (3\mathbf{k}).
\mathbf{u} \times \mathbf{v} = \langle 1, -4, 8 \rangle. Find \mathbf{v} \times \mathbf{u}, (-\mathbf{u}) \times (2\mathbf{v}) and the area of the triangle with sides \mathbf{u} and \mathbf{v}.
Find c so that \langle 2, c, 1 \rangle \times \langle 6, 9, 3 \rangle = \mathbf{0}.
Find both unit vectors perpendicular to \langle 1, 2, -1 \rangle and \langle 3, 0, 1 \rangle.
Find both vectors of length 14 perpendicular to \langle 3, -2, 0 \rangle and \langle 0, 2, -1 \rangle.

Points in space

Use a cross product to decide whether A(1, 2, 3), B(3, 5, 4) and C(7, 11, 6) lie on one line.
For the triangle of Exercise 6, find the height from R to the side PQ.

Reasoning

Show that \langle a, b, 0 \rangle \times \langle c, d, 0 \rangle = \langle 0, 0, ad - bc \rangle, and explain why |ad - bc| is the area of the parallelogram on \langle a, b \rangle and \langle c, d \rangle in the plane.
Show with \mathbf{i} and \mathbf{j} that \mathbf{u} \times \mathbf{v} and \mathbf{v} \times \mathbf{u} can differ, and explain from the 2 \times 2 blocks why every swap flips the sign.
Explain why knowing \|\mathbf{u}\|, \|\mathbf{v}\| and \|\mathbf{u} \times \mathbf{v}\| does not determine the angle between \mathbf{u} and \mathbf{v}, and what extra fact does.
Prove that \mathbf{u} \times \mathbf{v} is perpendicular to s\mathbf{u} + t\mathbf{v} for all numbers s and t.

Answers

\langle 3, -9, -12 \rangle. Components 3 \cdot 1 - (-2)(0) = 3, (-2)(4) - 1 \cdot 1 = -9, 1 \cdot 0 - 3 \cdot 4 = -12.
\langle 5, 15, -6 \rangle. With \langle 0, 2, 5 \rangle: 0 + 30 - 30 = 0. With \langle 3, -1, 0 \rangle: 15 - 15 + 0 = 0.
\langle 1, -1, 0 \rangle = \mathbf{i} - \mathbf{j}. (\mathbf{i} + \mathbf{j}) \times \mathbf{k} = \mathbf{i} \times \mathbf{k} + \mathbf{j} \times \mathbf{k} = -\mathbf{j} + \mathbf{i}.
\mathbf{0}. Components (-1)(-4) - 2 \cdot 2 = 0, 2(-6) - 3(-4) = 0, 3 \cdot 2 - (-1)(-6) = 0. The second vector is -2 times the first, so they are parallel.
\displaystyle \sqrt{101} \approx 10.05. The cross product is \langle 4, -7, -6 \rangle, and 16 + 49 + 36 = 101.
\displaystyle \sqrt{19} \approx 4.36. \overrightarrow{PQ} = \langle 2, 0, 2 \rangle, \overrightarrow{PR} = \langle 0, 3, 1 \rangle, cross product \langle -6, -2, 6 \rangle of length \displaystyle \sqrt{76} = 2\sqrt{19}; half of that.
12 and \displaystyle -12\sqrt{3}, from 24\sin 150^\circ and 24\cos 150^\circ. The length is positive even though the angle is obtuse.
\displaystyle \sin\theta = \frac{\sqrt{3}}{2}, and \theta = 60^\circ. The cross product \langle -1, -1, 1 \rangle has length \displaystyle \sqrt{3}, and both vectors have length \displaystyle \sqrt{2}. The dot product is 1 > 0, so the angle is the acute one, not 120^\circ.
C(5, 3, 5) and 20 square meters. C = B + \overrightarrow{AD} with \overrightarrow{AD} = \langle 1, 3, 4 \rangle. \langle 4, 0, 0 \rangle \times \langle 1, 3, 4 \rangle = \langle 0, -16, 12 \rangle, of length \displaystyle \sqrt{256 + 144} = 20.
14. \langle 6, 1, 0 \rangle \times \langle 2, 5, 0 \rangle = \langle 0, 0, 28 \rangle, and the triangle is half the parallelogram.
-\mathbf{k}, \mathbf{i} and -6\mathbf{j}. Backward around the cycle gives a negative; \mathbf{k} \times (-\mathbf{j}) = -(\mathbf{k} \times \mathbf{j}) = -(-\mathbf{i}); (2\mathbf{i}) \times (3\mathbf{k}) = 6(\mathbf{i} \times \mathbf{k}) = 6(-\mathbf{j}).
\langle -1, 4, -8 \rangle, \langle -2, 8, -16 \rangle and \displaystyle \frac{9}{2}. (-\mathbf{u}) \times (2\mathbf{v}) = -2(\mathbf{u} \times \mathbf{v}), and \displaystyle \|\langle 1, -4, 8 \rangle\| = \sqrt{81} = 9.
c = 3. The cross product is \langle 3c - 9, 0, 18 - 6c \rangle, zero only at c = 3, when \langle 6, 9, 3 \rangle = 3\langle 2, 3, 1 \rangle.
\displaystyle \pm\frac{1}{\sqrt{14}}\langle 1, -2, -3 \rangle. The cross product is \langle 2, -4, -6 \rangle = 2\langle 1, -2, -3 \rangle, and \displaystyle \|\langle 1, -2, -3 \rangle\| = \sqrt{14}.
\langle 4, 6, 12 \rangle and \langle -4, -6, -12 \rangle. The cross product is \langle 2, 3, 6 \rangle, of length 7; double it, or its negative.
Yes. \overrightarrow{AB} = \langle 2, 3, 1 \rangle and \overrightarrow{AC} = \langle 6, 9, 3 \rangle have cross product \mathbf{0}, so they are parallel; sharing the point A, they lie on one line.
\displaystyle \frac{\sqrt{38}}{2} \approx 3.08. The area \displaystyle \sqrt{19} is half the base \displaystyle \|\langle 2, 0, 2 \rangle\| = 2\sqrt{2} times the height, so the height is \displaystyle \frac{2\sqrt{19}}{2\sqrt{2}} = \sqrt{\frac{19}{2}}.
The first two components are b \cdot 0 - 0 \cdot d and 0 \cdot c - a \cdot 0, both 0; the third is ad - bc. The length |ad - bc| is the area of the parallelogram on the two vectors, which lie in the floor just as the plane vectors lie in the plane.
\mathbf{i} \times \mathbf{j} = \mathbf{k} but \mathbf{j} \times \mathbf{i} = -\mathbf{k}. Each component is a 2 \times 2 determinant with a row from the first vector above a row from the second; swapping the vectors swaps the rows, and cb - da = -(ad - bc).
They give only \sin\theta, and \sin\theta = \sin(180^\circ - \theta), so an angle and its supplement both fit (unless \theta = 90^\circ). The sign of \mathbf{u} \cdot \mathbf{v} decides: positive means acute, negative obtuse.
The dot product distributes and pulls out numbers, so (\mathbf{u} \times \mathbf{v}) \cdot (s\mathbf{u} + t\mathbf{v}) equals s\,(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{u} + t\,(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{v} = s \cdot 0 + t \cdot 0 = 0.