🤓 Based on our data, we think this question is relevant for Professor Leuenberger's class at UCF.

This problem asks for the scalar and vector products of two vectors, as well as the angle between them.

To finda vector's components from a magnitude and angle, we use these general equations for vectors in 3D:

$\overline{){{\mathit{A}}}_{{\mathit{x}}}{\mathbf{=}}\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\right|}\mathbf{}{\mathbf{cos}}{\mathbf{}}{\mathit{\alpha}}}$

$\overline{){{\mathit{A}}}_{{\mathit{y}}}{\mathbf{=}}\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\right|}\mathbf{}{\mathbf{cos}}{\mathbf{}}{\mathit{\beta}}}$

$\overline{){{\mathit{A}}}_{{\mathit{z}}}{\mathbf{=}}\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\right|}\mathbf{}{\mathbf{cos}}{\mathbf{}}{\mathit{\gamma}}}$ .

**Dot Product of Two Vectors.** Whenever we're asked to __find the dot product__ of two vectors, we'll use one of these two equations:

$\overline{)\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}{\mathbf{\xb7}}\stackrel{\mathbf{\rightharpoonup}}{\mathit{B}}{\mathbf{=}}{\mathit{A}}{\mathit{B}}{\mathbf{}}{\mathbf{cos}}{\mathbf{}}{\mathit{\theta}}}$ (the definition of dot product) or

$\overline{)\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}{\mathbf{\xb7}}\stackrel{\mathbf{\rightharpoonup}}{\mathit{B}}{\mathbf{=}}{{\mathit{A}}}_{{\mathit{x}}}{{\mathit{B}}}_{{\mathit{x}}}{\mathbf{+}}{{\mathit{A}}}_{{\mathit{y}}}{{\mathit{B}}}_{{\mathit{y}}}{\mathbf{+}}{{\mathit{A}}}_{{\mathit{z}}}{{\mathit{B}}}_{{\mathit{z}}}}$.

In most cases, using the component form (the second equation) is easiest, because it's simpler than finding *θ* if the vectors are 3D or given in components.

**Cross Product of Two Vectors.** To calculate the cross product of two vectors, it's almost always easiest to use the ** components**. You might remember that it's a long, complicated-looking equation. To make it easier to remember, we use this circle of unit-vectors and the directions they go with:

We'll show in part (b) how to use it!

**Angle Between Two Vectors.** When we have information about **both** the __dot product__ and the __cross product__ of two vectors, we can calculate the angle between them using the **definition of dot product** and the equation for the **magnitude of the cross product**, which is:

$\overline{)\mathbf{|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\times}\stackrel{\mathbf{\rightharpoonup}}{\mathit{B}}\mathbf{|}{\mathbf{=}}{\mathit{A}}{\mathit{B}}{\mathbf{}}{\mathbf{sin}}{\mathit{\theta}}}$

Dividing one equation by the other:

$\begin{array}{rcl}\frac{|\stackrel{\mathit{\rightharpoonup}}{\mathit{A}}\times \stackrel{\rightharpoonup}{\mathit{B}}|}{\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\xb7}\stackrel{\mathbf{\rightharpoonup}}{\mathit{B}}}& \mathbf{=}& \frac{\overline{)\mathit{A}\mathit{B}}\mathbf{}\mathbf{sin}\mathit{\theta}}{\overline{)\mathit{A}\mathit{B}}\mathbf{}\mathbf{cos}\mathit{\theta}}\\ \frac{|\stackrel{\rightharpoonup}{\mathit{A}}\times \stackrel{\mathit{\rightharpoonup}}{\mathit{B}}|}{\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\xb7}\stackrel{\mathbf{\rightharpoonup}}{\mathit{B}}}& \mathbf{=}& \mathbf{tan}\mathit{\theta}\mathbf{\u2007}\mathbf{\u2007}\mathbf{\u2007}\mathbf{\u2007}\mathbf{\u2007}\mathbf{\Rightarrow}\mathbf{\u2007}\overline{){\mathit{\theta}}{\mathbf{=}}{{\mathbf{tan}}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\times}\stackrel{\mathbf{\rightharpoonup}}{\mathit{B}}\mathbf{|}}{\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\xb7}\stackrel{\mathbf{\rightharpoonup}}{\mathit{B}}}}\end{array}$

If we have the cross product in unit-vector form, we calculate its magnitude using the equation

$\overline{){\mathbf{\left|}}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}{\mathbf{\right|}}{\mathbf{=}}\sqrt{{{\mathit{A}}_{\mathit{x}}}^{\mathbf{2}}\mathbf{+}{{\mathit{A}}_{\mathit{y}}}^{\mathbf{2}}\mathbf{+}{{\mathit{A}}_{\mathit{z}}}^{\mathbf{2}}}}$

Displacement **d**_{1} is in the *yz* plane 63.0° from the positive direction of the *y*-axis, has a positive *z* component, and has a magnitude of 4.50 m. Displacement **d**_{2} is in the *xz* plane 30.0° from the positive direction of the *x*-axis, has a positive *z* component, and has magnitude 1.40 m. What are

(a) **d**_{1} ⋅ **d**_{2}

(b) **d**_{1} ⨯ **d**_{2}, and

(c) the angle between **d**_{1} and **d**_{2}?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Intro to Cross Product (Vector Product) concept. If you need more Intro to Cross Product (Vector Product) practice, you can also practice Intro to Cross Product (Vector Product) practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Leuenberger's class at UCF.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Fundamentals of Physics - Halliday Calc 10th Edition. You can also practice Fundamentals of Physics - Halliday Calc 10th Edition practice problems.