Let a = (−1, 2, 2). Then find the directional cosines of a.

1 answer

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:

\[ \left( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3} \right). \]