Compute the dot product of the vectors \( u \) and \( v \), and find the angle between the vectors. \( u=\langle-12,0,5\rangle \) and \( \mathbf{v}=\langle 1,2,3\rangle \). \( \mathbf{u} \cdot \mathbf{v}=3 \). (Type an integer or a simplified fraction.) Find the magnitude of the vectors. \( |\mathbf{u}|=13 \) and \( |\mathbf{v}|=\sqrt{14} \) (Type exact answers, using radicals as needed.) The angle between the vectors is \( \square^{\circ} \). (Type your answer in degrees. Do not round until the final answer. Then round to the nearest hundredth as needed.)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Bonus Knowledge
To compute the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \), we use the formula: \[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \] Substituting the values: \[ \mathbf{u} \cdot \mathbf{v} = (-12)(1) + (0)(2) + (5)(3) = -12 + 0 + 15 = 3 \] Now, to find the angle \( \theta \) between the vectors, we can use the formula: \[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \] We already know \( \mathbf{u} \cdot \mathbf{v} = 3 \), \( |\mathbf{u}| = 13 \), and \( |\mathbf{v}| = \sqrt{14} \). Therefore, \[ \cos(\theta) = \frac{3}{13 \cdot \sqrt{14}} \] Now we can find \( \theta \): \[ \theta = \cos^{-1}\left(\frac{3}{13\sqrt{14}}\right) \] Using a calculator to evaluate \( \theta \): First, compute \( 13\sqrt{14} \): \[ 13\sqrt{14} \approx 13 \times 3.741657 = 48.632561 \] So, \[ \cos(\theta) \approx \frac{3}{48.632561} \approx 0.061674 \] Now take the inverse cosine: \[ \theta \approx \cos^{-1}(0.061674) \] Calculating this gives approximately: \[ \theta \approx 86.45^{\circ} \] Thus, the angle between the vectors is \( \boxed{86.45^{\circ}} \).