Question

Find the direction cosines of \(u\) and demonstrate that the sum of the squares of the direction cosines is 1. $$ \mathbf{u}=5 \mathbf{i}+3 \mathbf{j}-\mathbf{k} $$

Short answer

The direction cosines of the vector \(u = 5i + 3j -k\) are \(l = 5/\sqrt{35}\), \(m = 3/\sqrt{35}\), \(n = -1/\sqrt{35}\). The rule stating that the sum of the squares of direction cosines equals 1 has been verified.

Step-by-step solution

Calculate the magnitude of the vector

To calculate the direction cosines, we first need to know the magnitude of the vector. We calculate the magnitude (\(|u|\)) by using the formula: \[|u| = \sqrt{{u_x}^2+{u_y}^2+{u_z}^2}\] where \(u_x = 5, u_y = 3, u_z = -1\). This gives us \(|u| = \sqrt{5^2 +3^2+(-1)^2} = \sqrt{35}\).

Calculate the direction cosines

The direction cosines of a vector \(\mathbf{u}\) are the cosines of the angles that it makes with the positive x, y and z directions. The direction cosines are usually represented \(l, m, n\). To find direction cosines, we take the x, y and z components of the vector and divided by its magnitude. So we have \(l = \frac{u_x}{|u|} = 5/\sqrt{35}\), \(m = \frac{u_y}{|u|} = 3/\sqrt{35}\), \(n = \frac{u_z}{|u|} = -1/\sqrt{35}\).

Verify the sum of squares equals 1

Verify that the sum of the squares of the components equals to 1, as per the formula \(l^2 + m^2 + n^2 = 1\). Plug in \(l = 5/\sqrt{35}, m = 3/\sqrt{35}, n = -1/\sqrt{35}\) and simplify. We get \(\left(\frac{5}{\sqrt{35}}\right)^2 + \left(\frac{3}{\sqrt{35}}\right)^2 + \left(\frac{-1}{\sqrt{35}}\right)^2 = 1\). Hence, this verifies our results.