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# 5.6: Cross Products

Difficulty Level: At Grade Created by: CK-12
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Cross Products are related to Dot Products in a number of ways. Both are vector calculations, both are related not only to the magnitude of each vector, but also to the relative directions of both vectors.

Dot products in a sense calculate the joined force of two vectors in a given direction, and so are greatest when the vectors are parallel. Cross products, however, are greatest when the vectors are perpendicular... what then do they calculate?

### Watch This

'The video below is a well presented description of how to calculate cross-products. If you would like a deeper understanding of the difference between "Dot Products" and "Cross Products", take a look at: -Dot vs. Cross Product.'

Embedded Video:

### Guidance

Whereas a dot product of two vectors produces a scalar value; the cross product of the same two vectors produces a vector quantity having a direction perpendicular to the original two vectors.

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 is 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.

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

\begin{align*}\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\end{align*}

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

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 \begin{align*}+ \hat{z}\end{align*} or \begin{align*}-\hat{z}\end{align*}.

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, \begin{align*}\overrightarrow{A} \times \overrightarrow{B}\end{align*} is along \begin{align*}+\hat{z}\end{align*} (coming up out of the page) while \begin{align*}\overrightarrow{B} \times \overrightarrow{A}\end{align*} is along \begin{align*}-\hat{z}\end{align*} (going down into the page) and \begin{align*}\overrightarrow{A} \times \overrightarrow{B} = - \overrightarrow{B} \times \overrightarrow{A}\end{align*}

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 general, we can define a normal vector, \begin{align*}\hat{n}\end{align*}, 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.

\begin{align*}\hat{n} = \frac{\overrightarrow{U} \times \overrightarrow{V}} {|\overrightarrow{U} \times \overrightarrow{V}|}\end{align*}

#### Example A

Calculate the cross product of the two vectors shown below.

Solution

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

\begin{align*}\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\end{align*}

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.

\begin{align*}\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\end{align*}

\begin{align*}\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\end{align*}

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.

\begin{align*}\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 =\end{align*} \begin{align*}-4\end{align*}

Then find the magnitudes of the two vectors:

\begin{align*}| \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}\end{align*}

\begin{align*}| \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}\end{align*}

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

\begin{align*}\overrightarrow{A} \times \overrightarrow{B} = | A | | B |\ \mbox{cos}\ \theta\end{align*}

\begin{align*}-4 = \sqrt{15.25} \sqrt{20}\ \mbox{cos}\ \theta\end{align*}

\begin{align*}\mbox{cos}\ \theta = \frac{-4} {\sqrt{305}} \approx \frac{-4} {17.5} = -0.229\end{align*}

\begin{align*}\theta = 103^\circ\end{align*}

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

\begin{align*}| \overrightarrow{A} \times \overrightarrow{B} | = | \overrightarrow{A} | | \overrightarrow{B} | \ \mbox{sin}\ \theta\end{align*}

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

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 \begin{align*}+\hat{z}\end{align*} and \begin{align*}| \overrightarrow{A} \times \overrightarrow{B} | = \left \langle 0, 0, 17\right \rangle\end{align*}

#### Example B

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. \begin{align*}\overrightarrow{A} = \left \langle 3, 0, 4\right \rangle\end{align*} and \begin{align*}\overrightarrow{B} = \left \langle 5, 10, 0\right \rangle\end{align*}

Solution

The normal vector is defined by

\begin{align*}\hat{n} = \frac{\overrightarrow{U} \times \overrightarrow{V}} {|\overrightarrow{U} \times \overrightarrow{V}|}\end{align*}

In this case, we obtain

\begin{align*}\hat{n} = \frac{\overrightarrow{A} \times \overrightarrow{B}} {|\overrightarrow{A} \times \overrightarrow{B}|}\end{align*}

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

\begin{align*}\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\end{align*}

\begin{align*}\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\end{align*}

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

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

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

\begin{align*}\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\end{align*}

#### Example C

Determine the cross product \begin{align*}\overrightarrow{F} \times \overrightarrow{r}\end{align*} for the two vectors \begin{align*}\overrightarrow{F} = \left \langle 2, 3, 4\right \rangle\end{align*} and \begin{align*}\overrightarrow{r} = \left \langle 7, 6, 5\right \rangle\end{align*}. Then use the cross product to determine the angle between the two vectors.

Solution

One of the two ways to determine the magnitude of the cross product of two vectors uses the components of the two vectors:

\begin{align*}\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\end{align*}

\begin{align*}\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\end{align*}

\begin{align*}\overrightarrow{F} \times \overrightarrow{r} = \left \langle -9, 18, -9 \right \rangle\end{align*}

Now we can use the cross product and the second definition of the cross product to determine the angle between the two vectors.

\begin{align*}| \overrightarrow{F} \times \overrightarrow{r} | = | F | | r | \ \mbox{sin}\ \theta\end{align*}

We need to calculate the magnitudes of the vectors and of the cross product.

\begin{align*}| \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\end{align*}

\begin{align*}| \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} =\end{align*} \begin{align*}10.488\end{align*}

\begin{align*}| \overrightarrow{F} \times \overrightarrow{r}| = \sqrt{(-9)^2 + 18^2 + (-9)^2} = \sqrt{81 + 324 + 81} = \sqrt{486} = 22.0454\end{align*}

\begin{align*}\mbox{sin}\ \theta = \frac{|\overrightarrow{F} \times \overrightarrow{r}|} {| F | | r |} = \frac{22.0454} {(5.385)(10.488)} = 0.390\end{align*}

\begin{align*}\theta = \mbox{sin}^{-1} (0.390) = 22.98^\circ\end{align*}

We can use the dot product of the two vectors to check our solution.

\begin{align*}\overrightarrow{F} \times \overrightarrow{r} = |\overrightarrow{F}| |\overrightarrow{r}| \mbox{cos}\ \theta\end{align*}

\begin{align*}\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\end{align*}

\begin{align*}\mbox{cos}\ \theta = \frac{\overrightarrow{F} \times \overrightarrow{r}} {|\overrightarrow{F}| |\overrightarrow{r}|} = \frac{52} {(5.385)(10.488)} = 0.920714\end{align*}

\begin{align*}\theta = \mbox{cos}^{-1} (0.920714) = 22.97\end{align*}

This answer matches our value from the cross product to within rounding errors.

### Vocabulary

Cross products are a corollary to dot products, calculated as the area of a parallelogram with the two component vectors as the lengths of the sides.

The normal vector has a magnitude of one, and runs perpendicular to the plane formed by the two component vectors.

### Guided Practice

Questions

1) Determine the magnitude of the cross product of the two vectors shown below.

2) A plane passing through the origin is defined by the two vectors, \begin{align*}\overrightarrow{W} = \left \langle 4, 5, 2\right \rangle\end{align*} and \begin{align*}\overrightarrow{L} = \left \langle 8, 1, 9\right \rangle\end{align*}. Determine the equation of a unit vector representing a direction perpendicular to this plane.

3) Determine the area of a parallelogram whose sides are defined by the vectors \begin{align*}\overrightarrow{w} = \left \langle 85, 89, 91\right \rangle\end{align*} and \begin{align*}\overrightarrow{h} = \left \langle 67, 70, 88\right \rangle\end{align*}, lengths measured in centimeters.

4) Determine the cross product of the two vectors \begin{align*}\overrightarrow{f} = \left \langle 3, 13, 11\right \rangle\end{align*} and \begin{align*}\overrightarrow{g} = \left \langle 9, 6, 15\right \rangle\end{align*}

5) Determine the equation for the unit vector perpendicular to the plane defined by the two vectors \begin{align*}\overrightarrow{a} = \left \langle 2, 7, 4\right \rangle\end{align*} and \begin{align*}\overrightarrow{b} = \left \langle 0, 5, 1\right \rangle\end{align*}

6) Determine the area of the parallelogram whose sides are defined by \begin{align*}\overrightarrow{R} = \left \langle 27, 39, 52\right \rangle\end{align*} and \begin{align*}\overrightarrow{T} = \left \langle 44, 26, 17\right \rangle\end{align*}, lengths measured in millimeters.

7) Determine the magnitude of the cross-product of these two vectors.

Solutions

1) First we need to identify the components of the two vectors by using the information given on the graph. In this case, \begin{align*}\overrightarrow{MN} = \left \langle -2.25, 0.0\right \rangle\end{align*} and \begin{align*}\overrightarrow{KL} = \left \langle 1.5, 2, 0\right \rangle\end{align*}

\begin{align*}\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\end{align*}
\begin{align*}\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\end{align*}
\begin{align*}\overrightarrow{MN} \times \overrightarrow{KL} = \left \langle0 - 0, 0 - 0, -4.5 - 0\right \rangle = \left \langle0, 0, -4.5\right \rangle\end{align*}
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.
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2) To solve this problem we need to use the definition of the normal vector \begin{align*}\hat{n} = \frac{\overrightarrow{W} \times \overrightarrow{L}} {|\overrightarrow{W} \times \overrightarrow{L}|}\end{align*}, the component form of the definition of the cross product,

\begin{align*}\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\end{align*}. In this case, we obtain
\begin{align*}\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\end{align*}
\begin{align*}\overrightarrow{W} \times \overrightarrow{L} = \left \langle(45 - 2),(16 - 36), (4 - 40)\right \rangle = \left \langle43, -20, -36\right \rangle\end{align*}
We also need to know the magnitude of this cross product
\begin{align*}|\overrightarrow{W} \times \overrightarrow{L}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{(43)^2 + (-20)^2 + (-36)^2} = \sqrt{1849 + 400 + 1296} =\end{align*} \begin{align*}\sqrt{3545} = 59.54\end{align*}
Now we can determine the normal vector
\begin{align*}\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\end{align*}

3) 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, \begin{align*}|\overrightarrow{w} \times \overrightarrow{h}|\end{align*}. First we need to find the cross product of the two vectors:

\begin{align*}\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\end{align*}
\begin{align*}\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\end{align*}
\begin{align*}\overrightarrow{w} \times \overrightarrow{h} = \left \langle (7832 - 6370), (6097 - 7480), (5950 - 5963)\right \rangle = \left \langle 1462, -1383, -13\right \rangle\end{align*}
\begin{align*}| \overrightarrow{w} \times \overrightarrow{h} | = \sqrt{x^2 + y^2 + z^2} = \sqrt{1462^2 + (-1383)^2 + (13)^2} = \sqrt{4050302} \approx\end{align*}\begin{align*}2012.5\end{align*}
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.

4) \begin{align*}\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\end{align*}

\begin{align*}\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\end{align*}
\begin{align*}\overrightarrow{f} \times \overrightarrow{g} = \left \langle(195 - 66), (99 - 45), (18 - 117)\right \rangle = \left \langle 129, 54, -99\right \rangle\end{align*}

5) The cross product of two vectors is always perpendicular to the plane defined by the two vectors.

\begin{align*}\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\end{align*}
\begin{align*}\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\end{align*}
\begin{align*}\overrightarrow{a} \times \overrightarrow{b} = \left \langle(7 - 20), (0 - 2), (10 - 0)\right \rangle = \left \langle -13, -2, 10\right \rangle\end{align*}
The magnitude of this vector is given by
\begin{align*}| \overrightarrow{a} \times \overrightarrow{b}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{(-13)^2 + (-2)^2 + (10)^2} = \sqrt{273} = 16.5\end{align*}
Then divide the cross-product by its magnitude to obtain the unit vector.
\begin{align*}\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\end{align*}

6) 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, \begin{align*}|\overrightarrow{R} \times \overrightarrow{T}|\end{align*}. First we need to find the cross product of the two vectors:

\begin{align*}\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\end{align*}
\begin{align*}\overrightarrow{R} \times \overrightarrow{T} = \left \langle((39 * 17) - (52 * 26)), ((52 * 44) - (27 * 17)), ((27 * 26) - (39 * 44))\right \rangle\end{align*}
\begin{align*}\overrightarrow{R} \times \overrightarrow{T} = \left \langle((663) - (1352)), ((2288) - (459)), ((702) - (1716))\right \rangle\end{align*}
\begin{align*}\overrightarrow{R} \times \overrightarrow{T} = \left \langle(663 - 1352), (2288 - 459), (702 - 1716) \right \rangle = \left \langle -689, 1829, -1014\right \rangle\end{align*}
\begin{align*}|\overrightarrow{R} \times \overrightarrow{T}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{(-689)^2 + (1829)^2 + (-1014)^2} \approx 2202\end{align*}
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.
7) 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: \begin{align*}|\overrightarrow{A} \times \overrightarrow{B}| = | \overrightarrow{A}| | \overrightarrow{B}| \ \mbox{sin}\ \theta = (61)(45) \mbox{sin}\ 58 = 2328\end{align*}

Since these two vectors lie in the x-y plane, the direction of the cross-product will be parallel to the z-axis.

### Practice

Calculate the cross products:

1. Vectors c = \begin{align*}-6i + 2j + 3k\end{align*} and a = \begin{align*}-6i + 2j + 13k\end{align*}
2. Vectors v = \begin{align*}\left \langle-1,4,7\right \rangle\end{align*} and u = \begin{align*}\left \langle -5,10,3\right \rangle\end{align*}
3. Vectors f = \begin{align*}-6i + 8j + -6k\end{align*} and s = \begin{align*}-3i + 15j + 19k\end{align*}
4. Vectors j = \begin{align*}-3i + 15j + -4k\end{align*} and t = \begin{align*}7i + 10j + 6k\end{align*}
5. Vectors r = \begin{align*}\left \langle3,13,-1\right \rangle\end{align*} and v =\begin{align*} \left \langle7,6,1\right \rangle\end{align*}
6. Vectors e = \begin{align*}\left \langle-1,8,-3\right \rangle\end{align*} and a = \begin{align*}\left \langle -2,1,19\right \rangle\end{align*}
7. Vectors j = \begin{align*}-3i + 17j + 6k\end{align*} and h = \begin{align*}8i + 9j + 7k\end{align*}
8. Vectors a = \begin{align*}9i + 10j + 9k\end{align*} and g = \begin{align*}-5i + 19j + 15k\end{align*}
9. Vectors j = \begin{align*}4i + 18j + -8k\end{align*} and m = \begin{align*}2i + j + 19k\end{align*}
10. What is the cross product of \begin{align*}\left \langle-2,1,-2\right \rangle\end{align*} and \begin{align*}\left \langle5,6,9\right \rangle?\end{align*}
11. Find a vector orthogonal to both \begin{align*}\left \langle1,20,2\right \rangle\end{align*} and \begin{align*}\left \langle4,2,3\right \rangle\end{align*}
12. Vectors \begin{align*}y = 5i + 6j + 6k\end{align*} and \begin{align*}f = -4i + 9j + 3k.\end{align*} What is the area of the parallelogram formed by having y and f as adjacent sides?
13. What is the area of the parallelogram formed by having \begin{align*}\left \langle-2,1,7\right \rangle\end{align*} and \begin{align*}\left \langle5,7,16\right \rangle \end{align*} as adjacent sides?
14. What is the cross product between \begin{align*}\left \langle 8,6,8 \right \rangle\end{align*} and \begin{align*}\left \langle 8,6,8 \right \rangle\end{align*}?
15. Vectors \begin{align*} g = -6i + 9j + -7k\end{align*} and \begin{align*}y = -24i + 36j + -28k\end{align*} What is the cross product between g and y?
16. A boat is sailing on a bearing of 89° east of north at 564 feet per min. A tail wind is adding to the plane's velocity and blowing 78° west of north at 25 feet per min. Determine the actual speed of the plane in mph.
17. A plane is flying on a bearing of 77° east of south at 606 mph. A tail wind is adding to the plane's velocity and blowing 33° west of south at 80 mph. Determine the direction of the plane.

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### Vocabulary Language: English

Cross product

The cross product of two vectors is a third vector that is perpendicular to both of the original vectors.

dot product

The dot product is also known as inner product or scalar product. The two forms of the dot product are $\vec{a} \cdot \vec{b} = \Big \| \vec{a}\Big \| \ \Big \| \vec{b}\Big \| \cos \theta$ and $\vec{a} \cdot \vec{b} = x_a x_b + y_a y_b$.

normal vector

A normal vector is a vector that is perpendicular to a given surface or plane. A unit normal vector is a normal vector with a magnitude of one.

Right-hand-rule

The right-hand-rule is used to indicate the direction of a 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.

unit vector

A unit vector is a vector with a magnitude of one.

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Date Created:
Nov 01, 2012
Mar 23, 2016
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MAT.ALY.536.L.1
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