Question

Find the angle \(\theta\) between the vectors. $$ \begin{array}{l} \mathbf{u}=2 \mathbf{i}-3 \mathbf{j}+\mathbf{k} \\ \mathbf{v}=\mathbf{i}-2 \mathbf{j}+\mathbf{k} \end{array} $$

Short answer

The angle θ between the vectors u and v is approximately \(\arccos(\frac{9}{\sqrt{14} \sqrt{6}})\) degrees, which is about 16.26 degrees when evaluated.

Step-by-step solution

Calculating the Dot Product

Find the dot product of the vectors u and v. This is calculated by multiplying the corresponding i, j, k values between the two vectors and summing up those products. In this case, it would be \((2*1) + ((-3)*(-2)) + (1*1) = 2 + 6 + 1 = 9.\)

Calculating the Magnitude of Each Vector

The magnitude of a vector is calculated by squaring each i, j, k component, summing them all up, then taking the square root of that sum. For vector u, this would be \(\sqrt{(2^2) + ((-3)^2) + (1^2)} = \sqrt{14}\), and for vector v this would be \(\sqrt{(1^2) + ((-2)^2) + (1^2)} = \sqrt{6}\).

Calculating the Angle

The formula to calculate the angle between two vectors is \(\cos(\theta) = \frac{u \cdot v}{||u|| ||v||}\). Using the values calculated from the previous steps: \(\cos(\theta) = \frac{9}{\sqrt{14} \sqrt{6}}\). Now, to get theta, take the arccosine of the above value. Remember that your calculator should be in degree mode if you're looking to get the angle in degrees.