The Central Idea
Let
\[ \mathbf{u} = \langle u_1,u_2,u_3\rangle \]and
\[ \mathbf{v} = \langle v_1,v_2,v_3\rangle. \]Their cross product
\[ \boxed{ \mathbf{u}\times\mathbf{v} } \]is a vector perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\).
The Cross Product Produces a Vector
The cross product differs immediately from the dot product.
| Operation | Input | Output |
|---|---|---|
| Dot product | Two vectors | A scalar |
| Cross product | Two 3D vectors | A vector |
Why the Cross Product Is a Three-Dimensional Operation Here
In ordinary introductory vector algebra, the standard cross product is defined for vectors in three-dimensional space.
Two nonparallel vectors in three dimensions determine a plane.
The cross product identifies a direction perpendicular to that plane.
Component Formula
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
\[ \boxed{ \mathbf{u}\times\mathbf{v} = \left\langle u_2v_3-u_3v_2,\; u_3v_1-u_1v_3,\; u_1v_2-u_2v_1 \right\rangle }. \]A Useful Pattern
Each component of the cross product is built from two-by-two combinations of the original components.
The signs follow the pattern
\[ +,\;-,\;+ \]when the computation is remembered through determinant expansion.
Determinant-Style Memory Device
A common mnemonic writes
\[ \mathbf{u}\times\mathbf{v} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ u_1&u_2&u_3\\ v_1&v_2&v_3 \end{vmatrix}. \]Expanding formally along the first row gives
\[ \mathbf{i}(u_2v_3-u_3v_2) - \mathbf{j}(u_1v_3-u_3v_1) + \mathbf{k}(u_1v_2-u_2v_1). \]Why the Middle Component Often Causes Errors
The \(\mathbf{j}\)-term carries a minus sign in the determinant-style expansion:
\[ -\mathbf{j}(u_1v_3-u_3v_1). \]Distributing that negative gives
\[ (u_3v_1-u_1v_3)\mathbf{j}. \]This is why the component formula is often safer once it is familiar.
Example: Compute a Cross Product
Let
\[ \mathbf{u} = \langle 1,2,3\rangle \]and
\[ \mathbf{v} = \langle 4,5,6\rangle. \]Then
\[ \mathbf{u}\times\mathbf{v} = \left\langle 2(6)-3(5),\; 3(4)-1(6),\; 1(5)-2(4) \right\rangle. \]Complete the Example
Simplify:
\[ \mathbf{u}\times\mathbf{v} = \langle 12-15,\; 12-6,\; 5-8 \rangle. \]Therefore,
\[ \boxed{ \mathbf{u}\times\mathbf{v} = \langle -3,6,-3\rangle }. \]The Result Is Perpendicular to the First Vector
Let
\[ \mathbf{w} = \mathbf{u}\times\mathbf{v}. \]Then
\[ \boxed{ \mathbf{w}\cdot\mathbf{u}=0 }. \]Thus the cross product is orthogonal to \(\mathbf{u}\).
The Result Is Perpendicular to the Second Vector
The same cross product also satisfies
\[ \boxed{ \mathbf{w}\cdot\mathbf{v}=0 }. \]Therefore,
\[ \mathbf{u}\times\mathbf{v} \]is perpendicular to both original vectors.
Verify Perpendicularity in the Example
We found
\[ \mathbf{u}\times\mathbf{v} = \langle -3,6,-3\rangle. \]Dot with
\[ \mathbf{u}=\langle 1,2,3\rangle: \] \[ (-3)(1)+6(2)+(-3)(3) = -3+12-9 = 0. \]Verify against the Second Vector
Dot with
\[ \mathbf{v}=\langle 4,5,6\rangle: \] \[ (-3)(4)+6(5)+(-3)(6) = -12+30-18 = 0. \]The cross product is perpendicular to both vectors as required.
Two Possible Perpendicular Directions
A plane has two opposite normal directions.
If
\[ \mathbf{u}\times\mathbf{v} \]points in one normal direction, then
\[ -(\mathbf{u}\times\mathbf{v}) \]points in the opposite normal direction.
The order of the vectors determines which orientation the cross product chooses.
Order Matters
Cross products are not commutative.
Instead,
\[ \boxed{ \mathbf{v}\times\mathbf{u} = -(\mathbf{u}\times\mathbf{v}) }. \]Reversing the order reverses the direction of the resulting vector.
Example: Reverse the Order
If
\[ \mathbf{u}\times\mathbf{v} = \langle -3,6,-3\rangle, \]then
\[ \boxed{ \mathbf{v}\times\mathbf{u} = \langle 3,-6,3\rangle }. \]The magnitude is unchanged, but the direction is opposite.
The Right-Hand Rule
The right-hand rule gives the orientation of
\[ \mathbf{u}\times\mathbf{v}. \]Curl the fingers of your right hand from the direction of \(\mathbf{u}\) toward the direction of \(\mathbf{v}\) through the smaller angle.
Your thumb points in the direction of
\[ \mathbf{u}\times\mathbf{v}. \]Right-Handed Coordinate Orientation
For the standard basis vectors,
\[ \mathbf{i} = \langle 1,0,0\rangle, \qquad \mathbf{j} = \langle 0,1,0\rangle, \qquad \mathbf{k} = \langle 0,0,1\rangle, \]we have
\[ \boxed{ \mathbf{i}\times\mathbf{j} = \mathbf{k} }. \]The Cyclic Cross Products
In a right-handed coordinate system,
\[ \boxed{ \mathbf{i}\times\mathbf{j} = \mathbf{k} } \] \[ \boxed{ \mathbf{j}\times\mathbf{k} = \mathbf{i} } \] \[ \boxed{ \mathbf{k}\times\mathbf{i} = \mathbf{j} }. \]These follow the cyclic order
\[ \mathbf{i}\to\mathbf{j}\to\mathbf{k}\to\mathbf{i}. \]Reverse the Cyclic Order
Reversing the order introduces a negative sign:
\[ \boxed{ \mathbf{j}\times\mathbf{i} = -\mathbf{k} } \] \[ \boxed{ \mathbf{k}\times\mathbf{j} = -\mathbf{i} } \] \[ \boxed{ \mathbf{i}\times\mathbf{k} = -\mathbf{j} }. \]A Vector Crossed with Itself
For every vector \(\mathbf{v}\),
\[ \boxed{ \mathbf{v}\times\mathbf{v} = \mathbf{0} }. \]A vector is parallel to itself, so there is no nonzero perpendicular area generated by the pair.
Parallel Vectors Have Zero Cross Product
If nonzero vectors \(\mathbf{u}\) and \(\mathbf{v}\) are parallel, then
\[ \boxed{ \mathbf{u}\times\mathbf{v} = \mathbf{0} }. \]Their cross product has zero magnitude because the vectors do not span a genuine parallelogram.
A Cross-Product Test for Parallelism
For nonzero vectors in three dimensions,
\[ \boxed{ \mathbf{u}\parallel\mathbf{v} \iff \mathbf{u}\times\mathbf{v} = \mathbf{0} }. \]This complements the dot-product test for perpendicularity.
Geometric Magnitude Formula
If \(\theta\) is the angle between nonzero vectors \(\mathbf{u}\) and \(\mathbf{v}\), then
\[ \boxed{ \|\mathbf{u}\times\mathbf{v}\| = \|\mathbf{u}\|\, \|\mathbf{v}\| \sin\theta }. \]This gives the geometric size of the cross product.
Why Sine Appears
Imagine \(\mathbf{u}\) as the base of a parallelogram.
The component of \(\mathbf{v}\) perpendicular to \(\mathbf{u}\) has length
\[ \|\mathbf{v}\|\sin\theta. \]Therefore,
\[ \text{base}\times\text{height} = \|\mathbf{u}\|\|\mathbf{v}\|\sin\theta. \]Cross Product Magnitude as Parallelogram Area
The area of the parallelogram generated by \(\mathbf{u}\) and \(\mathbf{v}\) is
\[ \boxed{ A_{\text{parallelogram}} = \|\mathbf{u}\times\mathbf{v}\| }. \]This is one of the most important geometric interpretations of the cross product.
Triangle Area from a Cross Product
A triangle formed by the same two side vectors has half the parallelogram area.
Therefore,
\[ \boxed{ A_{\triangle} = \frac12 \|\mathbf{u}\times\mathbf{v}\| }. \]Example: Area of a Parallelogram
Let
\[ \mathbf{u} = \langle 2,0,0\rangle \]and
\[ \mathbf{v} = \langle 0,3,0\rangle. \]Their cross product is
\[ \mathbf{u}\times\mathbf{v} = \langle 0,0,6\rangle. \]Thus the parallelogram area is
\[ \boxed{6}. \]Triangle Area in the Same Example
Since
\[ \|\mathbf{u}\times\mathbf{v}\|=6, \]the triangle area is
\[ \boxed{ \frac12(6)=3 }. \]Area from Three Points
Suppose a triangle has vertices \(A\), \(B\), and \(C\).
Two side vectors from the same vertex are
\[ \overrightarrow{AB} \]and
\[ \overrightarrow{AC}. \]Therefore,
\[ \boxed{ A_{\triangle ABC} = \frac12 \left\| \overrightarrow{AB} \times \overrightarrow{AC} \right\| }. \]Example: Triangle from Coordinates
Let
\[ A=(0,0,0), \qquad B=(2,0,0), \qquad C=(0,3,0). \]Then
\[ \overrightarrow{AB} = \langle 2,0,0\rangle \]and
\[ \overrightarrow{AC} = \langle 0,3,0\rangle. \]The triangle area is \(3\).
Finding a Normal Vector to a Plane
If two nonparallel vectors
\[ \mathbf{u} \]and
\[ \mathbf{v} \]lie in a plane, then
\[ \boxed{ \mathbf{u}\times\mathbf{v} } \]is a normal vector to that plane.
Why the Cross Product Gives a Plane Normal
A normal vector must be perpendicular to every direction generated by the two spanning vectors.
Since
\[ (\mathbf{u}\times\mathbf{v})\cdot\mathbf{u}=0 \]and
\[ (\mathbf{u}\times\mathbf{v})\cdot\mathbf{v}=0, \]the cross product is perpendicular to both directions and therefore to the plane they span.
Example: Find a Plane Normal
Let the plane contain
\[ \mathbf{u} = \langle 1,2,0\rangle \]and
\[ \mathbf{v} = \langle 0,1,3\rangle. \]Then
\[ \mathbf{u}\times\mathbf{v} = \left\langle 2(3)-0(1),\; 0(0)-1(3),\; 1(1)-2(0) \right\rangle. \]Complete the Plane-Normal Example
Therefore,
\[ \boxed{ \mathbf{u}\times\mathbf{v} = \langle 6,-3,1\rangle }. \]Any nonzero scalar multiple of this vector is also a normal vector to the same plane.
Normal Direction Is Not Unique
If
\[ \mathbf{n} = \mathbf{u}\times\mathbf{v}, \]then
\[ -\mathbf{n} \]is also perpendicular to the plane.
A plane has two opposite normal orientations.
Cross Product and the Standard Basis
The basis cross products can be remembered using the cycle
\[ \mathbf{i} \to \mathbf{j} \to \mathbf{k} \to \mathbf{i}. \]Moving forward in the cycle gives a positive basis vector.
Moving backward gives its negative.
Cross Product of Perpendicular Unit Vectors
If \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular unit vectors, then
\[ \|\mathbf{u}\|=\|\mathbf{v}\|=1 \]and
\[ \theta=90^\circ. \]Therefore,
\[ \|\mathbf{u}\times\mathbf{v}\| = 1(1)\sin90^\circ = \boxed{1}. \]Cross Product of Parallel Unit Vectors
If the unit vectors point in the same direction, then
\[ \theta=0^\circ. \]Therefore,
\[ \|\mathbf{u}\times\mathbf{v}\| = \sin0^\circ = 0. \]Their cross product is the zero vector.
Cross Product of Opposite Vectors
If two nonzero vectors point in opposite directions, then
\[ \theta=180^\circ. \]Since
\[ \sin180^\circ=0, \]their cross product is also the zero vector.
The Dot Product Uses Cosine; the Cross Product Uses Sine
The geometric formulas are
\[ \mathbf{u}\cdot\mathbf{v} = \|\mathbf{u}\|\|\mathbf{v}\|\cos\theta \]and
\[ \|\mathbf{u}\times\mathbf{v}\| = \|\mathbf{u}\|\|\mathbf{v}\|\sin\theta. \]The dot product measures parallel alignment, while cross-product magnitude measures perpendicular spread.
Parallel and Perpendicular Information
For nonzero vectors:
- \(\mathbf{u}\cdot\mathbf{v}=0\) indicates perpendicularity;
- \(\mathbf{u}\times\mathbf{v}=\mathbf{0}\) indicates parallelism.
The two operations reveal complementary geometric relationships.
Cross Product and Angle
If the magnitudes are known, cross-product magnitude can also be used to find the sine of the angle:
\[ \boxed{ \sin\theta = \frac{ \|\mathbf{u}\times\mathbf{v}\| }{ \|\mathbf{u}\|\|\mathbf{v}\| } }. \]However, sine alone generally does not distinguish an acute angle from its supplementary obtuse angle.
Dot and Cross Products Together
Using both operations gives stronger angular information:
\[ \mathbf{u}\cdot\mathbf{v} = \|\mathbf{u}\|\|\mathbf{v}\|\cos\theta \]and
\[ \|\mathbf{u}\times\mathbf{v}\| = \|\mathbf{u}\|\|\mathbf{v}\|\sin\theta. \]Together, cosine and sine determine the angle more completely.
Distributive Property
The cross product distributes over vector addition:
\[ \boxed{ \mathbf{u}\times(\mathbf{v}+\mathbf{w}) = \mathbf{u}\times\mathbf{v} + \mathbf{u}\times\mathbf{w} }. \]Similarly,
\[ (\mathbf{u}+\mathbf{v})\times\mathbf{w} = \mathbf{u}\times\mathbf{w} + \mathbf{v}\times\mathbf{w}. \]Scalars and the Cross Product
For a scalar \(k\),
\[ \boxed{ (k\mathbf{u})\times\mathbf{v} = k(\mathbf{u}\times\mathbf{v}) }. \]Also,
\[ \mathbf{u}\times(k\mathbf{v}) = k(\mathbf{u}\times\mathbf{v}). \]Cross Product Is Anticommutative
Instead of ordinary commutativity,
\[ \boxed{ \mathbf{u}\times\mathbf{v} = -\mathbf{v}\times\mathbf{u} }. \]This property is called anticommutativity.
The Cross Product Is Not Associative
In general,
\[ \boxed{ (\mathbf{u}\times\mathbf{v})\times\mathbf{w} \neq \mathbf{u}\times(\mathbf{v}\times\mathbf{w}) }. \]Parentheses therefore matter in expressions containing repeated cross products.
Cross Product with the Zero Vector
For every vector \(\mathbf{v}\),
\[ \boxed{ \mathbf{v}\times\mathbf{0} = \mathbf{0} }. \]Likewise,
\[ \mathbf{0}\times\mathbf{v} = \mathbf{0}. \]Finding a Unit Normal
A cross product gives a normal vector, but not necessarily one of magnitude \(1\).
If
\[ \mathbf{n} = \mathbf{u}\times\mathbf{v} \neq \mathbf{0}, \]then a unit normal is
\[ \boxed{ \widehat{\mathbf{n}} = \frac{\mathbf{n}}{\|\mathbf{n}\|} }. \]Two Unit Normals
If
\[ \widehat{\mathbf{n}} \]is one unit normal, then
\[ -\widehat{\mathbf{n}} \]is the opposite unit normal.
Choosing between them depends on the orientation required by the problem.
Example: Find a Unit Normal
Suppose
\[ \mathbf{u}\times\mathbf{v} = \langle 0,3,4\rangle. \]Its magnitude is
\[ \sqrt{0^2+3^2+4^2} = 5. \]Therefore, one unit normal is
\[ \boxed{ \left\langle 0,\frac35,\frac45 \right\rangle }. \]Torque as a Cross-Product Idea
In mechanics, torque can be modeled by
\[ \boxed{ \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} }. \]The magnitude measures the turning effect of the force, while the direction indicates the axis and orientation of rotation.
Why Perpendicular Force Produces Maximum Turning Effect
Since
\[ \|\mathbf{r}\times\mathbf{F}\| = \|\mathbf{r}\|\|\mathbf{F}\|\sin\theta, \]the magnitude is largest when
\[ \theta=90^\circ. \]A force parallel to \(\mathbf{r}\) gives zero cross product and no turning effect in this idealized model.
Angular Momentum as Another Cross-Product Idea
In mechanics, angular momentum may be written
\[ \boxed{ \mathbf{L} = \mathbf{r}\times\mathbf{p} }. \]Again, the cross product combines magnitude, perpendicular separation, and orientation.
Cross Products and Plane Equations
Suppose three noncollinear points
\[ A,\quad B,\quad C \]determine a plane.
Compute
\[ \mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}. \]Then \(\mathbf{n}\) is a normal vector that can be used in a point-normal plane equation.
From Three Points to a Plane Normal
If
\[ A=(x_1,y_1,z_1), \] \[ B=(x_2,y_2,z_2), \] \[ C=(x_3,y_3,z_3), \]first form
\[ \overrightarrow{AB} = \langle x_2-x_1,\; y_2-y_1,\; z_2-z_1 \rangle \]and
\[ \overrightarrow{AC} = \langle x_3-x_1,\; y_3-y_1,\; z_3-z_1 \rangle. \]Then Take the Cross Product
Compute
\[ \boxed{ \mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} }. \]Provided the points are noncollinear, this cross product is nonzero and normal to the plane.
Collinear Points Produce a Zero Cross Product
If \(A\), \(B\), and \(C\) are collinear, then
\[ \overrightarrow{AB} \]and
\[ \overrightarrow{AC} \]are parallel.
Therefore,
\[ \boxed{ \overrightarrow{AB} \times \overrightarrow{AC} = \mathbf{0} }. \]Three collinear points do not determine a unique plane.
Common Error: Returning a Scalar
A cross product returns a vector.
The expression
\[ \mathbf{u}\times\mathbf{v} \]should produce three components in the standard three-dimensional setting.
A scalar result usually indicates confusion with the dot product.
Common Error: Forgetting the Middle Sign
In determinant-style expansion, the signs are
\[ +,\;-,\;+. \]Forgetting the negative on the middle cofactor changes the direction of the result and usually destroys perpendicularity.
Common Error: Reversing the Order
In general,
\[ \mathbf{u}\times\mathbf{v} \neq \mathbf{v}\times\mathbf{u}. \]Instead,
\[ \boxed{ \mathbf{v}\times\mathbf{u} = -(\mathbf{u}\times\mathbf{v}) }. \]Always preserve the stated order.
Common Error: Using the Left Hand for Orientation
Standard cross-product orientation uses the right-hand rule.
Using the opposite hand reverses the resulting normal direction.
Common Error: Confusing Zero Cross Product with Perpendicularity
A zero dot product indicates perpendicular nonzero vectors.
A zero cross product indicates parallel nonzero vectors.
These two tests should not be interchanged.
Common Error: Taking the Magnitude Too Early
The cross product itself is a vector:
\[ \mathbf{u}\times\mathbf{v}. \]Its magnitude
\[ \|\mathbf{u}\times\mathbf{v}\| \]is a scalar.
Use the magnitude only when the problem asks for area, size, or a scalar quantity.
Common Error: Forgetting the Factor of \(\frac12\) for Triangle Area
The magnitude
\[ \|\mathbf{u}\times\mathbf{v}\| \]gives parallelogram area.
Triangle area is
\[ \boxed{ \frac12 \|\mathbf{u}\times\mathbf{v}\| }. \]Common Error: Assuming the Cross Product Is Associative
Expressions such as
\[ (\mathbf{u}\times\mathbf{v})\times\mathbf{w} \]and
\[ \mathbf{u}\times(\mathbf{v}\times\mathbf{w}) \]generally differ.
Keep parentheses exactly as specified.
A Reliable Procedure for Computing a Cross Product
- Write both vectors in three-component form.
- Preserve the order \(\mathbf{u}\times\mathbf{v}\).
- Compute the first component \(u_2v_3-u_3v_2\).
- Compute the second component \(u_3v_1-u_1v_3\).
- Compute the third component \(u_1v_2-u_2v_1\).
- Check signs carefully.
- Verify perpendicularity with dot products when useful.
A Reliable Procedure for Finding a Plane Normal
- Identify two nonparallel vectors lying in the plane.
- Compute their cross product.
- Check that the result is nonzero.
- Verify that its dot product with each plane direction is zero.
- Simplify by a common scalar factor if convenient.
A Reliable Procedure for Area
- Identify two side vectors sharing a common initial point.
- Compute their cross product.
- Find the magnitude of the cross product.
- Use that magnitude directly for parallelogram area.
- Multiply by \(\frac12\) for triangle area.
A Reliable Procedure for Cross Products from Three Points
- Choose one point as the common initial point.
- Form two displacement vectors from that point to the other two points.
- Take their cross product.
- Use the result as a normal vector or area vector as required.
- If the cross product is zero, recognize that the three points are collinear.
Example: Full Cross-Product Workflow
Let
\[ \mathbf{u} = \langle 2,-1,3\rangle \]and
\[ \mathbf{v} = \langle 1,4,-2\rangle. \]Compute
\[ \mathbf{u}\times\mathbf{v}. \]Compute the First Component
\[ u_2v_3-u_3v_2 = (-1)(-2)-3(4). \]Therefore,
\[ 2-12=-10. \]Compute the Second Component
\[ u_3v_1-u_1v_3 = 3(1)-2(-2). \]Therefore,
\[ 3+4=7. \]Compute the Third Component
\[ u_1v_2-u_2v_1 = 2(4)-(-1)(1). \]Therefore,
\[ 8+1=9. \]Complete the Full Cross-Product Workflow
Combining the components,
\[ \boxed{ \mathbf{u}\times\mathbf{v} = \langle -10,7,9\rangle }. \]Verify the Full Example
Dot with \(\mathbf{u}\):
\[ (-10)(2)+7(-1)+9(3) = -20-7+27 = 0. \]Dot with \(\mathbf{v}\):
\[ (-10)(1)+7(4)+9(-2) = -10+28-18 = 0. \]The result is perpendicular to both original vectors.
Practice 1: Compute a Cross Product
Find
\[ \langle 1,0,0\rangle \times \langle 0,1,0\rangle. \]Practice 2: Reverse the Order
Find
\[ \langle 0,1,0\rangle \times \langle 1,0,0\rangle. \]Practice 3: General Computation
Find
\[ \langle 2,1,0\rangle \times \langle 0,3,4\rangle. \]Practice 4: Parallelism Test
Determine whether
\[ \mathbf{u} = \langle 1,2,3\rangle \]and
\[ \mathbf{v} = \langle 2,4,6\rangle \]are parallel using the cross product.
Practice 5: Parallelogram Area
Find the area of the parallelogram generated by
\[ \mathbf{u} = \langle 3,0,0\rangle \]and
\[ \mathbf{v} = \langle 0,4,0\rangle. \]Practice 6: Triangle Area
Using the same vectors from Practice 5, find the area of the triangle they generate.
Practice 7: Find a Normal Vector
Find a normal vector to a plane containing
\[ \mathbf{u} = \langle 1,0,2\rangle \]and
\[ \mathbf{v} = \langle 0,3,1\rangle. \]Practice 8: Unit Normal
If a normal vector is
\[ \mathbf{n} = \langle 0,6,8\rangle, \]find one unit normal in the same direction.
Practice 9: Three-Point Area
Let
\[ A=(0,0,0), \qquad B=(1,0,0), \qquad C=(0,2,0). \]Find the area of triangle \(ABC\).
Practice 10: Identify the Error
A student writes
\[ \mathbf{u}\times\mathbf{v} = \mathbf{v}\times\mathbf{u} \]because multiplication is commutative.
Explain the error.
Practice Answers
Practice 1
\[ \boxed{ \langle 1,0,0\rangle \times \langle 0,1,0\rangle = \langle 0,0,1\rangle }. \]Thus,
\[ \mathbf{i}\times\mathbf{j} = \mathbf{k}. \]Practice 2
\[ \boxed{ \langle 0,1,0\rangle \times \langle 1,0,0\rangle = \langle 0,0,-1\rangle }. \]Reversing the order reverses the direction.
Practice 3
\[ \langle 2,1,0\rangle \times \langle 0,3,4\rangle = \langle 1(4)-0(3),\; 0(0)-2(4),\; 2(3)-1(0) \rangle. \]Therefore,
\[ \boxed{ \langle 4,-8,6\rangle }. \]Practice 4
Compute:
\[ \langle 1,2,3\rangle \times \langle 2,4,6\rangle = \langle 0,0,0\rangle. \]Therefore, the nonzero vectors are parallel.
Practice 5
\[ \mathbf{u}\times\mathbf{v} = \langle 0,0,12\rangle. \]Therefore,
\[ \boxed{ A_{\text{parallelogram}}=12 }. \]Practice 6
\[ \boxed{ A_{\triangle} = \frac12(12) = 6 }. \]Practice 7
\[ \langle 1,0,2\rangle \times \langle 0,3,1\rangle = \langle 0(1)-2(3),\; 2(0)-1(1),\; 1(3)-0(0) \rangle. \]Therefore, one normal vector is
\[ \boxed{ \langle -6,-1,3\rangle }. \]Practice 8
The magnitude is
\[ \sqrt{0^2+6^2+8^2}=10. \]Therefore, one unit normal is
\[ \boxed{ \left\langle 0,\frac35,\frac45 \right\rangle }. \]Practice 9
\[ \overrightarrow{AB} = \langle 1,0,0\rangle \]and
\[ \overrightarrow{AC} = \langle 0,2,0\rangle. \]Their cross product has magnitude \(2\).
Therefore,
\[ \boxed{ A_{\triangle ABC} = \frac12(2) = 1 }. \]Practice 10
The cross product is not commutative.
It is anticommutative:
\[ \boxed{ \mathbf{u}\times\mathbf{v} = -(\mathbf{v}\times\mathbf{u}) }. \]Reversing the order reverses the normal orientation.
What to Remember
- The standard cross product combines two 3D vectors and returns a vector.
- \(\mathbf{u}\times\mathbf{v}\) is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\).
- Cross-product order matters: \(\mathbf{v}\times\mathbf{u}=-(\mathbf{u}\times\mathbf{v})\).
- The right-hand rule determines the orientation of the cross product.
- \(\|\mathbf{u}\times\mathbf{v}\|=\|\mathbf{u}\|\|\mathbf{v}\|\sin\theta\).
- Cross-product magnitude equals the area of the generated parallelogram.
- Triangle area is half the magnitude of the corresponding cross product.
- Parallel nonzero vectors have zero cross product.
- A cross product of two nonparallel in-plane vectors gives a plane normal.
- The cross product is distributive but not commutative or associative.
Where We Go Next
The cross product gives us a perpendicular direction to two vectors and therefore a natural normal vector to a plane. This makes it especially useful when moving from vector operations into coordinate descriptions of geometric objects.
The next lesson, Vector Equations of Lines and Planes, uses points, direction vectors, spanning vectors, dot products, and normal vectors to describe lines and planes algebraically.