Question

Find the direction cosines of the vector \(3\vec{i} - 4\vec{j} + 12\vec{k}\).

Hint:

Direction cosines of a vector are the cosines of the angles the vector makes with the coordinate axes and are found by dividing each component by the vector's magnitude.

Answer 1

First, find the magnitude of the vector: \[ |\vec{v}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13. \] The direction cosines are given by: \[ \cos \alpha = \frac{3}{13}, \quad \cos \beta = \frac{-4}{13}, \quad \cos \gamma = \frac{12}{13}. \]
Final answer: \[ \boxed{ \left( \frac{3}{13}, \; -\frac{4}{13}, \; \frac{12}{13} \right). } \]