Question

In Exercises 29 and 30 . find the direction angles of the vector. $$ \mathbf{u}=3 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k} $$

Short answer

The direction angles of the vector \( \mathbf{u} \) are \( \cos^{-1}(3/\sqrt{17}) \), \( \cos^{-1}(2/\sqrt{17}) \), and \( \cos^{-1}(-2/\sqrt{17}) \) respectively.

Step-by-step solution

Calculate the Magnitude of the Vector

First, calculate the magnitude of the vector by using the formula \( \|\mathbf{u}\|= \sqrt{a^2 + b^2 + c^2} \), where a, b, and c are the x, y, and z-components of the vector respectively. So, \( \|\mathbf{u}\| = \sqrt{3^2+2^2+(-2)^2} = \sqrt{17} \).

Calculate the Direction Angles

Next, calculate the direction angles \(\alpha\), \(\beta\), and \(\gamma\) which are the angles the vector makes with the positive x, y, and z axes respectively. Use the formulas \(\cos(\alpha) = a / \|\mathbf{u}\|\), \(\cos(\beta) = b / \|\mathbf{u}\|\), \(\cos(\gamma) = c / \|\mathbf{u}\|\). Plugging our vector components and the magnitude in, we get \(\cos(\alpha) = 3 / \sqrt{17}\), \(\cos(\beta) = 2 / \sqrt{17}\), \(\cos(\gamma) = -2 / \sqrt{17}\). These give us the direction cosines, from which we can find the direction angles by taking the inverse cosine of each value.