Question

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

Answer 1

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:

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