Question

Find the direction cosines of \(u\) and demonstrate that the sum of the squares of the direction cosines is 1. $$ \mathbf{u}=\langle a, b, c\rangle $$

Short answer

The direction cosines of vector \( \mathbf{u} = \langle a, b, c \rangle \) are \( l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} \). And the sum of the squares of the direction cosines is indeed equal to 1, as per the direction cosines property.

Step-by-step solution

Find the magnitude of vector \( \mathbf{u} \)

The magnitude of vector \( \mathbf{u} \) is given by the formula \( ||\mathbf{u}|| = \sqrt{a^2 + b^2 + c^2} \).

Compute the direction cosines

The direction cosines, denoted as \( l, m, n \) respectively, which are the cosines of the angles that vector \( \mathbf{u} \) makes with the positive x, y and z axes. They are computed as follows: \[ l = \frac{a}{||\mathbf{u}||}, m = \frac{b}{||\mathbf{u}||}, n = \frac{c}{||\mathbf{u}||} \].

Demonstrate that the sum of squares is 1

The demonstration involves squaring each of the computed direction cosines and adding them together. The result should be equal to one. This is done as follows: \[ l^2 + m^2 + n^2 = \left(\frac{a}{||\mathbf{u}||}\right)^2 + \left(\frac{b}{||\mathbf{u}||}\right)^2 + \left(\frac{c}{||\mathbf{u}||}\right)^2 = \frac{a^2 + b^2 + c^2 }{||\mathbf{u}||^2} = 1 \]