6. $||\vec{u}|| = 10$, $||\vec{w}|| = 20$; the angle between $\vec{u}$ and $\vec{w}$ is $120^\circ$.
Added by Kimberly O.
Close
Step 1
Step 1: The dot product of two vectors is given by: $\vec{u} \cdot \vec{w} = ||\vec{u}|| ||\vec{w}|| \cos \theta$ Show more…
Show all steps
Your feedback will help us improve your experience
Kamalesh Kumar and 64 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Evaluate $\vec{u} \cdot \vec{w}$ $\|\vec{u}\|=10,\|\vec{w}\|=20 ;$ the angle between $\vec{u}$ and $\vec{w}$ is $120^{\circ}$
A Fundamental Tool: Vectors
The Dot Product
Evaluate $\vec{u} \cdot \vec{w}$ $\|\vec{u}\|=3,\|\vec{w}\|=5 ;$ the angle between $\vec{u}$ and $\vec{w}$ is $45^{\circ}$
In each part use the given information to find u v. $$ \begin{array}{l}{\text { (a) }\|\mathbf{u}\|=1,\|\mathbf{v}\|=2, \text { the angle between } \mathbf{u} \text { and } \mathbf{v} \text { is } \pi / 6} \\ {\text { (b) }\|\mathbf{u}\|=2,\|\mathbf{v}\|=3, \text { the angle between } \mathbf{u} \text { and } \mathbf{v} \text { is } 135^{\circ} .}\end{array} $$
THREE-DIMENSIONAL SPACE; VECTORS
Dot Product; Projections
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD