Question
Part 2 of 3 Compute the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \), and find the angle between the vectors. \( \mathbf{u}=\langle-12,0,5\rangle \) and \( \mathbf{v}=\langle 1,2,3\rangle \). \( \mathbf{u} \cdot \mathbf{v}=3 \). (Type an integer or a simplified fraction.) 1 Find the magnitude of the vectors. \( |\mathbf{u}|=\square \) and \( |\mathbf{v}|=\square \) (Type exact answers, using radicals as needed.)
Ask by Summers Harmon. in the United States
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The magnitudes of the vectors are:
\[
|\mathbf{u}| = 13 \quad \text{and} \quad |\mathbf{v}| = \sqrt{14}
\]
Solution
To find the magnitudes of the vectors \( \mathbf{u} \) and \( \mathbf{v} \), we'll use the formula for the magnitude of a vector:
\[
|\mathbf{w}| = \sqrt{w_1^2 + w_2^2 + w_3^2}
\]
### For \( \mathbf{u} = \langle -12, 0, 5 \rangle \):
\[
|\mathbf{u}| = \sqrt{(-12)^2 + 0^2 + 5^2} = \sqrt{144 + 0 + 25} = \sqrt{169} = 13
\]
### For \( \mathbf{v} = \langle 1, 2, 3 \rangle \):
\[
|\mathbf{v}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}
\]
### **Final Answers:**
\[
|\mathbf{u}| = 13 \quad \text{and} \quad |\mathbf{v}| = \sqrt{14}
\]
Reviewed and approved by the UpStudy tutoring team
Like
error msg
Explain 
Simplify this solution Extra Insights
To find the magnitudes of the vectors \( \mathbf{u} \) and \( \mathbf{v} \): 1. The magnitude of vector \( \mathbf{u} = \langle -12, 0, 5 \rangle \) is calculated using the formula \( |\mathbf{u}| = \sqrt{(-12)^2 + 0^2 + 5^2} \). \[ |\mathbf{u}| = \sqrt{144 + 0 + 25} = \sqrt{169} = 13 \] 2. The magnitude of vector \( \mathbf{v} = \langle 1, 2, 3 \rangle \) is calculated similarly: \[ |\mathbf{v}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \] So, the final answers are: \( |\mathbf{u}| = 13 \) and \( |\mathbf{v}| = \sqrt{14} \).
