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5.4: Cross Products

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Learning objectives

  • Calculate the cross product of a pair of vectors.
  • Apply the Right Hand Rule to correctly identify the direction of the vector product.
  • Use the cross product to identify the direction perpendicular to a plane.

Introduction

In the last section we discussed the dot product of two vectors. In this section we will discuss the other form of vector multiplication. The dot product of two vectors produced a scalar value; the cross product of the same two vectors will produce a vector quantity having a direction perpendicular to the original two vectors.

Cross Product

The cross product of two vector quantities is another vector whose magnitude varies as the angle between the two original vectors changes. The cross product therefore sometimes referred to as the vector product of two vectors. The magnitude of the cross product represents the area of the parallelogram whose sides are defined by the two vectors, as shown in the figure below. Therefore, the maximum value for the cross product occurs when the two vectors are perpendicular to one another, but when the two vectors are parallel to one another the magnitude of the cross product is equal to zero.

Like the dot product, the cross product must satisfy several important properties of multiplication. The cross product is distributive,

\overrightarrow{A} \times (\overrightarrow{B} + \overrightarrow{C}) = \overrightarrow{A} \times \overrightarrow{B} + \overrightarrow{A} \times \overrightarrow{C}

But, unlike the dot product, the cross product is anti-commutative

\overrightarrow{A} \times \overrightarrow{B} = -\overrightarrow{B} \times \overrightarrow{A}

Computing the Cross Product

The algebraic form of the cross product equation is more complicated than that for the dot product. For two 3D vectors \overrightarrow{A} and \overrightarrow{B},

\overrightarrow{A} \times \overrightarrow{B} = \left \langle (A_2 B_3 - A_3 B_2),(A_3 B_1 - A_1 B_3), (A_1 B_2 - A_2 B_1)\right \rangle

Another way to describe the process is to say that the cross product is the multiplication of one vector by the component of the other vector which is perpendicular to the first vector. In the diagram below are two vectors, A and B. A perpendicular line has been drawn radially outward from B towards A to create a right triangle with A as the hypotenuse.

The component of \overrightarrow{A} which is perpendicular to \overrightarrow{B} is given by A sin θ so the magnitude of the cross product can be written as | \overrightarrow{A} \times \overrightarrow{B} | = \overrightarrow{A} (\overrightarrow{B} \mbox{sin}\ \theta) = | \overrightarrow{A} | | \overrightarrow{B} | \ \mbox{sin}\ \theta

Likewise, the component of \overrightarrow{B} which is parallel to \overrightarrow{A} is given by B sin θ, so the cross product | \overrightarrow{B} \times \overrightarrow{A} | = | \overrightarrow{B} | (| \overrightarrow{A} |  \ \mbox{sin}\ \theta ) = | \overrightarrow{A} | | \overrightarrow{B} | \ \mbox{sin}\ \theta

The value of the cross product is maximized when the two vectors are perpendicular and zero when the two vectors are parallel to one another. When a vector is crossed with itself, the result is the zero vector since a vector has no component perpendicular to itself.

| \overrightarrow{A} \times \overrightarrow{A} | = | \overrightarrow{A} | (| \overrightarrow{A} | \ \mbox{sin}\ \theta ) = | \overrightarrow{A} | | \overrightarrow{A} | \ \mbox{sin}\ 0 = 0

Also, the cross product of any vector with the zero vector is equal to zero since the magnitude of the zero vector is itself equal to 0.

We mentioned earlier, that the direction of the cross product is perpendicular to the plane defined by the two crossed vectors. For example, the cross product of two vectors in the x-y plane will be parallel to the z-axis. This still leaves two possible directions for the cross product, though: either + \hat{z} or -\hat{z}.

We use a right-hand-rule to indicate the direction of the cross product. Position the thumb and index finger of your right hand with the first vector along your thumb and the second vector along your index finger. Your middle finger, when extended perpendicular to your palm, will indicate the direction of the cross product of the two vectors.

As you can see in the diagram above, \overrightarrow{A} \times \overrightarrow{B} is along +\hat{z} (coming up out of the page) while \overrightarrow{B} \times \overrightarrow{A} is along -\hat{z} (going down into the page) and \overrightarrow{A} \times \overrightarrow{B} = - \overrightarrow{B} \times \overrightarrow{A}

Example: Calculate the cross product of the two vectors shown below.

Solution: Use the components of the two vectors to determine the cross product.

\overrightarrow{A} \times \overrightarrow{B} = \left \langle (A_y B_z - A_z B_y), (A_zB_x - A_x B_z), (A_x B_y - A_y B_x)\right \rangle

Since these two vectors are both in the x-y plane, their own z-components are both equal to 0 and the vector product will be parallel to the z axis.

\overrightarrow{A} \times \overrightarrow{B} = \left \langle [(3 \cdot 0) - (0 \cdot 2)], [(0 \cdot -4) - (2.5 \cdot 0)], [(2.5 \cdot 2) - (3 \cdot -4)]\right \rangle

\overrightarrow{A} \times \overrightarrow{B} = \left \langle [(0) - (0)], [(0) - (0)], [(5) - (-12)]\right \rangle = \left \langle 0,0,(5 + 12)\right \rangle = \left \langle 0,0,17\right \rangle

We can check our answer using the sine version of the cross product, but first we need to know the angle between the two vectors. We can use the dot product to find θ, following the procedure in the first Example in the previous section. First use the components to find the dot product.

\overrightarrow{A} \times \overrightarrow{B} = A_x B_x + A_y B_y + A_z B_z = (2.5 * -4) + (3 * 2) + (0 * 0) = -10 + 6 + 0 = -4

Then find the magnitudes of the two vectors:

| \overrightarrow{A} | = \sqrt{A^2_x + A^2_y + A^2_z} = \sqrt{2.5^2 + 3^2 + 0^2} = \sqrt{6.25 + 9 + 0} = \sqrt{15.25}

| \overrightarrow{B} | = \sqrt{B^2_x + B^2_y + B^2_z} = \sqrt{(-4)^2 + 2^2 + 0^2} = \sqrt{16 + 4 + 0} = \sqrt{20}

Then use these magnitudes with the cosine version of the dot product to find θ.

\overrightarrow{A} \times \overrightarrow{B} = | A | | B |\ \mbox{cos}\ \theta

-4 = \sqrt{15.25} \sqrt{20}\ \mbox{cos}\ \theta

\mbox{cos}\ \theta = \frac{-4} {\sqrt{305}} \approx \frac{-4} {17.5} = -0.229

\theta = 103^\circ

Now use the sine of this angle and the two magnitudes to determine the cross product:

| \overrightarrow{A} \times \overrightarrow{B} | = | \overrightarrow{A} | | \overrightarrow{B} | \ \mbox{sin}\ \theta

| \overrightarrow{A} \times \overrightarrow{B} | = \sqrt{15.25} \sqrt{20}\ \mbox{sin}\ 103^\circ = \sqrt{305}\ \mbox{sin}\ 103^\circ = 17

This is the same answer that we obtained from the component notation, which is good. We use the RHR to determine the direction of the vector product. If you place your thumb along vector A and your forefinger along vector B, your middle finger will point along +\hat{z} and | \overrightarrow{A} \times \overrightarrow{B} | = \left \langle 0, 0, 17\right \rangle

The Normal Vector

We can use the cross product and the definition of the unit vector to determine the direction which is perpendicular to a plane. In the previous example, the cross product of the two vectors had a magnitude given by

| \overrightarrow{A} \times \overrightarrow{B} | = \sqrt{15.25}\ \sqrt{20}\ \mbox{sin}\ 103^\circ = \sqrt{305}\ \mbox{sin}\ 103^\circ = 17

and we used the right hand rule to show that it pointed along +\hat{z} such that

\overrightarrow{A} \times \overrightarrow{B} = \left \langle0, 0, 17\right \rangle

In general, we can define a normal vector, \hat{n}, which has a unity magnitude (i.e. magnitude equal to one) and which is perpendicular to a plane occupied by a pair of vectors, U and V.

\hat{n} = \frac{\overrightarrow{U} \times \overrightarrow{V}} {|\overrightarrow{U} \times \overrightarrow{V}|}

For the vectors A and B in the previous problem

\hat{n} = \frac{\overrightarrow{A} \times \overrightarrow{B}} {|\overrightarrow{A} \times \overrightarrow{B}|} = \frac{\left \langle 0, 0, 17 \right \rangle} {17} = \left \langle \frac{0} {17}, \frac{0} {17}, \frac{17} {17} \right \rangle = \left \langle 0, 0, 1\right \rangle

which is in the + \hat{z} direction, as shown above using the right hand rule.

Example: The diagram shows two vectors A and B which define a plane passing through the origin. Use these two vectors to determine the normal vector to this plane. \overrightarrow{A} = \left \langle 3, 0, 4\right \rangle and \overrightarrow{B} = \left \langle 5, 10, 0\right \rangle

Solution: The normal vector is defined by

\hat{n} = \frac{\overrightarrow{U} \times \overrightarrow{V}} {|\overrightarrow{U} \times \overrightarrow{V}|}

In this case, we obtain

\hat{n} = \frac{\overrightarrow{A} \times \overrightarrow{B}} {|\overrightarrow{A} \times \overrightarrow{B}|}

Use the component version of the cross-product equation to find the components of \overrightarrow{A} \times \overrightarrow{B}

\overrightarrow{A} \times \overrightarrow{B} = \left \langle (A_y B_z - A_z B_y), (A_z B_x - A_x B_z), (A_x B_y - A_y B_x)\right \rangle

\overrightarrow{A} \times \overrightarrow{B} = \left \langle [(0 \cdot 0) - (4 \cdot 10)], [(4 \cdot 5) - (3 \cdot 0)], [(3 \cdot 10) - (0 \cdot 5)]\right \rangle

\overrightarrow{A} \times \overrightarrow{B} = \left \langle (0 - 40), (20 - 0), (30 - 0)\right \rangle = \left \langle -40, 20, 30\right \rangle

Next, calculate the magnitude of the cross product, | \overrightarrow{A} \times \overrightarrow{B}|

| \overrightarrow{A} \times \overrightarrow{B}| = \sqrt{(-40)^2 + 20^2 + 30^2} = \sqrt{1600 + 400 + 900} = \sqrt {2900} = 53.8516

\hat{n} = \frac{\overrightarrow{A} \times \overrightarrow{B}} {|\overrightarrow{A} \times \overrightarrow{B}|} = \frac{\left \langle -40, 20, 30\right \rangle} {53.9} = \left \langle \frac{-40} {53.9}, \frac{20} {53.9}, \frac{30} {53.9}\right \rangle = \left \langle -0.743, 0.371, 0.557\right \rangle

Lesson Summary

One of the two ways to multiply vector quantities is the Vector Product. The vector product, also known as the cross product, multiplies one vector by the component of the second vector which is perpendicular to the first. The result of a scalar product of two vectors is always a vector quantity which is perpendicular to the plane defined by the first two vectors. There are two ways to calculate the dot product: \overrightarrow{A} \times \overrightarrow{B} = \left \langle(A_y B_z - A_z B_y), (A_z B_x - A_x B_z), (A_x B_y - A_y B_x)\right \rangle and the magnitude of the cross product is given by | \overrightarrow{A} \times \overrightarrow{B}| = | A | | B | \ \mbox{sin}\ \theta. These two versions of the cross product can be use to determine the angle between two vectors. The cross product can also be used to identify the direction perpendicular to a plane.

Practice Problems

  1. Determine the magnitude and direction of the cross product \overrightarrow{F} \times \overrightarrow{r} for the two vectors \overrightarrow{F} = \left \langle 2, 3, 4\right \rangle and \overrightarrow{r} = \left \langle 7, 6, 5\right \rangle. Then use the cross product to determine the angle between the two vectors.
  2. Determine the magnitude of the cross product of the two vectors shown below.
  3. A plane passing through the origin is defined by the two vectors, \overrightarrow{W} = \left \langle 4, 5, 2\right \rangle and \overrightarrow{L} = \left \langle 8, 1, 9\right \rangle. Determine the equation of a unit vector representing a direction perpendicular to this plane.
  4. Determine the area of a parallelogram whose sides are defined by the vectors \overrightarrow{w} = \left \langle 85, 89, 91\right \rangle and \overrightarrow{h} = \left \langle 67, 70, 88\right \rangle, lengths measured in centimeters.
  5. Determine the magnitude of the cross product of the two vectors \overrightarrow{f} = \left \langle 3, 13, 11\right \rangle and \overrightarrow{g} = \left \langle 9, 6, 15\right \rangle.
  6. Determine the equation for the unit vector perpendicular to the plane defined by the two vectors \overrightarrow{a} = \left \langle 2, 7, 4\right \rangle and \overrightarrow{b} = \left \langle 0, 5, 1\right \rangle.
  7. Determine the area of the parallelogram whose sides are defined by \overrightarrow{R} = \left \langle 27, 39, 52\right \rangle and \overrightarrow{T} = \left \langle 44, 26, 17\right \rangle, lengths measured in millimeters.
  8. Determine the magnitude and direction of the cross-product of these two vectors.

Solutions

  1. Determine the cross product \overrightarrow{F} \times \overrightarrow{r} for the two vectors \overrightarrow{F} = \left \langle 2, 3, 4\right \rangle and \overrightarrow{r} = \left \langle 7, 6, 5\right \rangle. Then use the cross product to determine the angle between the two vectors. One of the two ways to determine the magnitude of the cross product of two vectors uses the components of the two vectors: \overrightarrow{F} \times \overrightarrow{r} = \left \langle (F_y r_z - F_z r_y), (F_z r_x - F_x r_z),(F_x r_y - F_y r_x)\right \rangle \overrightarrow{F} \times \overrightarrow{r} = \left \langle(3 \cdot 5 - 4 \cdot 6), (4 \cdot 7 - 2 \cdot 5), (2 \cdot 6 - 3 \cdot 7)\right \rangle = \left \langle(15 - 24), (28 - 10), (12 - 21)\right \rangle \overrightarrow{F} \times \overrightarrow{r} = \left \langle -9, 18, -9 \right \rangle Now we can use the cross product and the second definition of the cross product to determine the angle between the two vectors. | \overrightarrow{F} \times \overrightarrow{r} | = | F | | r | \ \mbox{sin}\ \theta We need to calculate the magnitudes of the vectors and of the cross product. | \overrightarrow{F} | = \sqrt{F^2_x + F^2_y + F^2_z} = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{ 4 + 9 + 16} = \sqrt{29} = 5.385 | \overrightarrow{r}| = \sqrt{r^2_x + r^2_y + r^2_z} = \sqrt{7^2 + 6^2 + 5^2} = \sqrt{49 + 36 + 25} = \sqrt{110} = 10.488 | \overrightarrow{F} \times \overrightarrow{r}| = \sqrt{(-9)^2 + 18^2 + (-9)^2} = \sqrt{81 + 324 + 81} = \sqrt{486} = 22.0454 \mbox{sin}\ \theta = \frac{|\overrightarrow{F} \times \overrightarrow{r}|} {| F |  | r |} = \frac{22.0454} {(5.385)(10.488)} = 0.390 \theta = \mbox{sin}^{-1} (0.390) = 22.98^\circ We can use the dot product of the two vectors to check our solution. \overrightarrow{F} \times \overrightarrow{r} = |\overrightarrow{F}| |\overrightarrow{r}| \mbox{cos}\ \theta \overrightarrow{F} \times \overrightarrow{r} = F_x r_x + F_y r_y + F_z r_z = 2*7 + 3*6 + 4*5 = 14 + 18 + 20 = 52 \mbox{cos}\ \theta = \frac{\overrightarrow{F} \times \overrightarrow{r}} {|\overrightarrow{F}| |\overrightarrow{r}|} = \frac{52} {(5.385)(10.488)} = 0.920714 \theta = \mbox{cos}^{-1} (0.920714) = 22.97 This answer matches our value from the cross product to within rounding errors.
  2. Determine the magnitude of the cross product of the two vectors shown below. First we need to identify the components of the two vectors by using the information given on the graph. In this case, \overrightarrow{MN} = \left \langle -2.25, 0.0\right \rangle and \overrightarrow{KL} = \left \langle 1.5, 2, 0\right \rangle. \overrightarrow{MN} \times \overrightarrow{KL} = \left \langle(MN_y KL_z- MN_z KL_y), (MN_z KL_x - MN_x KL_z), (MN_x KL_y - MN_y KL_x)\right \rangle \overrightarrow{MN} \times \overrightarrow{KL} = \left \langle(0 \cdot 0 - 0 \cdot 2), (0 \cdot 1.5 - (-2.25) \cdot 0), ((-2.25) \cdot 2 - 0 \cdot 1.5)\right \rangle \overrightarrow{MN} \times \overrightarrow{KL} = \left \langle0 - 0, 0 - 0, -4.5 - 0\right \rangle = \left \langle0, 0, -4.5\right \rangle As we can see by the components, this vector has a magnitude of 4.5 units and lies in the –z direction. We can also use the Right Hand Rule to see the direction of the cross product. As shown in the figure below, if we align the right thumb with vector MN and the right fore-finger with vector KL, the palm and extended middle-finger point in the –z direction.
  3. A plane passing through the origin is defined by the two vectors, \overrightarrow{W} = \left \langle 4, 5, 2\right \rangle and \overrightarrow{L} = \left \langle 8, 1, 9\right \rangle. Determine the equation of a unit vector representing a direction perpendicular to this plane. To solve this problem we need to use the definition of the normal vector \hat{n} = \frac{\overrightarrow{W} \times \overrightarrow{L}} {|\overrightarrow{W} \times \overrightarrow{L}|}, the component form of the definition of the cross product, \overrightarrow{W} \times \overrightarrow{L} = \left \langle(W_y L_z - W_z L_y), (W_z L_x - W_x L_z), (W_x L_y - W_y L_x)\right \rangle. In this case, we obtain \overrightarrow{W} \times \overrightarrow{L} = \left \langle(5 \cdot 9 - 2 \cdot 1),(2 \cdot 8 - 4 \cdot 9), (4 \cdot 1 - 5 \cdot 8)\right \rangle \overrightarrow{W} \times \overrightarrow{L} = \left \langle(45 - 2),(16 - 36), (4 - 40)\right \rangle = \left \langle43, -20, -36\right \rangle We also need to know the magnitude of this cross product |\overrightarrow{W} \times \overrightarrow{L}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{(43)^2 + (-20)^2 + (-36)^2} = \sqrt{1849 + 400 + 1296} = \sqrt{3545} = 59.54 Now we can determine the normal vector \hat{n} = \frac{\overrightarrow{W} \times \overrightarrow{L}} {| \overrightarrow{W} \times \overrightarrow{L}|} = \frac{\left \langle 43, -20, -36\right \rangle} {59.54} = \left \langle \frac{43} {59.54}, \frac{-20} {59.54}, \frac{-36} {59.54}\right \rangle = \left \langle 0.7222, -0.3359, -0.6046\right \rangle
  4. Determine the area of a parallelogram whose sides are defined by the vectors \overrightarrow{w} = \left \langle 85, 89, 91\right \rangle and \overrightarrow{h} = \left \langle 67, 70, 88\right \rangle, lengths measured in centimeters. The area of the parallelogram whose sides are defined by a pair of vectors is equal to the magnitude of the cross product of the two vectors, |\overrightarrow{w} \times \overrightarrow{h}|. First we need to find the cross product of the two vectors: \overrightarrow{w} \times \overrightarrow{h} = \left \langle(w_y h_z - w_z h_y), (w_z h_x - w_x h_z), (w_x h_y - w_y h_x)\right \rangle \overrightarrow{w} \times \overrightarrow{h} = \left \langle (89 \cdot 88 - 91 \cdot 70), (91 \cdot 67 - 85 \cdot 88), (85 \cdot 70 - 89 \cdot 67)\right \rangle \overrightarrow{w} \times \overrightarrow{h} = \left \langle (7832 - 6370), (6097 - 7480), (5950 - 5963)\right \rangle = \left \langle 1462, -1383, -13\right \rangle | \overrightarrow{w} \times \overrightarrow{h} | = \sqrt{x^2 + y^2 + z^2} = \sqrt{1462^2 + (-1383)^2 + (13)^2} = \sqrt{4050302} \approx 2012.5 Since the lengths of the two vectors were measured in centimeters, the area of the parallelogram is 2013 cm2 measured to the nearest square centimeter.
  5. Determine the cross product of the two vectors \overrightarrow{f} = \left \langle 3, 13, 11\right \rangle and \overrightarrow{g} = \left \langle 9, 6, 15\right \rangle. \overrightarrow{f} \times \overrightarrow{g} = \left \langle(f_y g_z - f_z g_y), (f_z g_x - f_x g_z), (f_x g_y - f_y g_x)\right \rangle \overrightarrow{f} \times \overrightarrow{g} = \left \langle(13 \cdot 15 - 11 \cdot 6), (11 \cdot 9 - 3 \cdot 15), (3 \cdot 6 - 13 \cdot 9)\right \rangle \overrightarrow{f} \times \overrightarrow{g} = \left \langle(195 - 66), (99 - 45), (18 - 117)\right \rangle = \left \langle 129, 54, -99\right \rangle
  6. Determine the equation for the unit vector perpendicular to the plane defined by the two vectors \overrightarrow{a} = \left \langle 2, 7, 4\right \rangle and \overrightarrow{b} = \left \langle 0, 5, 1\right \rangle. The cross product of two vectors is always perpendicular to the plane defined by the two vectors. \overrightarrow{a} \times \overrightarrow{b} = \left \langle(a_y b_z - a_z b_y), (a_z b_x - a_x b_z), (a_x b_y - a_y b_x)\right \rangle \overrightarrow{a} \times \overrightarrow{b} = \left \langle((7 \cdot 1) - (4 \cdot 5)), ((4 \cdot 0) - (2 \cdot 1)), ((2 \cdot 5) - (7 \cdot 0))\right \rangle \overrightarrow{a} \times \overrightarrow{b} = \left \langle(7 - 20), (0 - 2), (10 - 0)\right \rangle = \left \langle -13, -2, 10\right \rangle The magnitude of this vector is given by | \overrightarrow{a} \times \overrightarrow{b}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{(-13)^2 + (-2)^2 + (10)^2} = \sqrt{273} = 16.5 Then divide the cross-product by its magnitude to obtain the unit vector. \hat{n} = \frac{\overrightarrow{a} \times \overrightarrow{b}} {|\overrightarrow{a} \times \overrightarrow{b}|} = \frac{\left \langle -13, -2, 10\right \rangle} {16.5} = \left \langle \frac{-13} {16.5}, \frac{-2} {16.5}, \frac{10} {16.5}\right \rangle
  7. Determine the area of the parallelogram whose sides are defined by \overrightarrow{R} = \left \langle 27, 39, 52\right \rangle and \overrightarrow{T} = \left \langle 44, 26, 17\right \rangle, lengths measured in millimeters. The area of the parallelogram whose sides are defined by a pair of vectors is equal to the magnitude of the cross product of the two vectors, |\overrightarrow{R} \times \overrightarrow{T}|. First we need to find the cross product of the two vectors: \overrightarrow{R} \times \overrightarrow{T} = \left \langle(R_y T_z - R_z T_y), (R_z T_x - R_x T_z), (R_x T_y - R_y T_x)\right \rangle \overrightarrow{R} \times \overrightarrow{T} = \left \langle((39 * 17) - (52 * 26)), ((52 * 44) - (27 * 17)), ((27 * 26) - (39 * 44))\right \rangle \overrightarrow{R} \times \overrightarrow{T} = \left \langle((663) - (1352)), ((2288) - (459)), ((702) - (1716))\right \rangle \overrightarrow{R} \times \overrightarrow{T} = \left \langle(663 - 1352), (2288 - 459), (702 - 1716) \right \rangle = \left \langle -689, 1829, -1014\right \rangle |\overrightarrow{R} \times \overrightarrow{T}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{(-689)^2 + (1829)^2 + (-1014)^2} \approx 2202 Since the lengths of the two vectors were measured in centimeters, the area of the parallelogram is 2202 mm2 measured to the nearest square centimeter.
  8. Determine the magnitude of the cross-product of these two vectors. Since we know the magnitudes of the two vectors and the angle between them, we can use the angle-version of the cross-product equation to determine the magnitude of the cross-product: |\overrightarrow{A} \times \overrightarrow{B}| = | \overrightarrow{A}| | \overrightarrow{B}| \ \mbox{sin}\ \theta = (61)(45) \mbox{sin}\ 58 = 2328 Since these two vectors lie in the x-y plane, the direction of the cross-product will be parallel to the z-axis.

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Feb 23, 2012

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Sep 24, 2014
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