Question

Find the angle \(\theta\) between the vectors. $$ \begin{array}{l} \mathbf{u}=\langle 1,1,1\rangle \\ \mathbf{v}=\langle 2,1,-1\rangle \end{array} $$

Short answer

\(\theta = \cos^{-1}(\frac{\sqrt{2}}{3}) \approx 0.955\) radians or 54.7 degrees

Step-by-step solution

Calculate the Dot Product

First, the dot product of the two vectors is computed. The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is calculated as \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \). Substituting the values of the vectors, \( \mathbf{u} \cdot \mathbf{v} = (1)(2) + (1)(1) + (1)(-1) = 2

Compute the Magnitudes of the Vectors

Next, compute the magnitudes of the vectors. The magnitude of a vector \( \mathbf{u}=\langle u_1,u_2,u_3 \rangle \) is calculated as \( ||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2 + u_3^2} \). So, \( ||\mathbf{u}|| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \) and \( ||\mathbf{v}|| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6} \)

Find the Cosine of the Angle

The cosine of the angle \( \theta \) between the vectors is calculated as \( \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| ||\mathbf{v}||} \). Substituting the calculated values, \( \cos(\theta) = \frac{2}{\sqrt{3} \sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{\sqrt{2}}{3} \)

Calculate the Angle

Lastly, the angle \( \theta \) is calculated by taking the inverse cosine of \( \frac{\sqrt{2}}{3} \). Thus, \( \theta = \cos^{-1}(\frac{\sqrt{2}}{3}) \), which is approximately 0.955 radians or 54.7 degrees.