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 compute the magnitude (length) of the vector.

The magnitude \( ||\mathbf{a}|| \) is calculated using the formula:

\[
||\mathbf{a}|| = \sqrt{(-1)^2 + 2^2 + 2^2}
\]

Calculating this gives:

\[
||\mathbf{a}|| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
\]

The directional cosines are defined as the cosines of the angles that the vector makes with the coordinate axes (x, y, z).

The directional cosines \( l, m, n \) corresponding to the vector \( \mathbf{a} \) are given by:

\[
l = \frac{a_x}{||\mathbf{a}||}, \quad m = \frac{a_y}{||\mathbf{a}||}, \quad n = \frac{a_z}{||\mathbf{a}||}
\]

Where \( a_x, a_y, a_z \) are the components of the vector \( \mathbf{a} \). For our vector:

- \( a_x = -1 \)
- \( a_y = 2 \)
- \( a_z = 2 \)

Now, we can calculate each of the directional cosines:

\[
l = \frac{-1}{3}, \quad m = \frac{2}{3}, \quad n = \frac{2}{3}
\]

Therefore, the directional cosines of the vector \( \mathbf{a} \) are:

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