Question

In Exercises \(25-28,\) find the direction cosines of \(u\) and demonstrate that the sum of the squares of the direction cosines is 1. $$ \mathbf{u}=\mathbf{i}+2 \mathbf{j}+2 \mathbf{k} $$

Short answer

The direction cosines of the vector \(\mathbf{u}=\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}\) are \(\cos \alpha = \frac{1}{3}, \cos \beta = \frac{2}{3}, \cos \gamma = \frac{2}{3}\). The sum of the squares of these direction cosines equals 1, which validates Pythagorean theorem in three dimensions.

Step-by-step solution

Find the magnitude of the vector

The magnitude of the vector is calculated using the formula \(|u| = \sqrt{(x^2 + y^2 + z^2)}\) where \(x, y, z\) are the coefficients of \(\mathbf{i, j, k}\) respectively. For the given vector \(\mathbf{u}=\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}\), we have \(x=1, y=2, z =2\). So, \(|u| = \sqrt{(1^2 + 2^2 + 2^2)} = \sqrt{9} = 3.\)

Calculate the direction cosines

The direction cosines \(\cos \alpha, \cos \beta, \cos \gamma\) are the ratios of the vector's coefficients to its magnitude and correspond to the angles \(\alpha, \beta, \gamma\) that the vector makes with the x, y, and z axes respectively. They can be calculated as follows: \[\cos \alpha = \frac{x}{|u|}, \cos \beta = \frac{y}{|u|}, \cos \gamma = \frac{z}{|u|}\]So, we have:\[\cos \alpha = \frac{1}{3}, \cos \beta = \frac{2}{3}, \cos \gamma = \frac{2}{3}\]

Verify that the squares of the direction cosines sum to 1

The sum of the squares of the direction cosines should equal 1. We verify this as follows:\[\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \left(\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{1}{9} + \frac{4}{9} + \frac{4}{9} = \frac{9}{9} = 1\]So, the sum of the squares of the direction cosines does indeed equal 1, as required.