To find the directional cosines of the vector \( \mathbf{a} = (-1, 2, 2) \), we first need to calculate the magnitude (length) of the vector:
\[ |\mathbf{a}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3. \]
Next, we can find the directional cosines, which are the cosines of the angles that the vector makes with the coordinate axes. The directional cosines are given by:
\[ l = \frac{a_x}{|\mathbf{a}|}, \quad m = \frac{a_y}{|\mathbf{a}|}, \quad n = \frac{a_z}{|\mathbf{a}|}, \]
where \( a_x, a_y, \) and \( a_z \) are the components of the vector \( \mathbf{a} \).
In our case:
- \( a_x = -1 \)
- \( a_y = 2 \)
- \( a_z = 2 \)
Now we can compute the directional cosines:
\[ l = \frac{-1}{3}, \quad m = \frac{2}{3}, \quad n = \frac{2}{3}. \]
So, the directional cosines of the vector \( \mathbf{a} = (-1, 2, 2) \) are:
\[ \left( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3} \right). \]
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