Question

Explain why it is not possible for two of a vector's direction angles to be less than 45°

Answer 1

Answer on Question #40758 – Math – Vector Calculus

Explain why it is not possible for two of a vector's direction angles to be less than 45.

Solution.

As it is known, there is a theorem for vector's direction angles, which states that

$$\cos^2 \alpha_x + \cos^2 \alpha_y + \cos^2 \alpha_z = 1.$$

This is statement can be easily derived, if one substitutes $\frac{a_x}{a}$ for $\alpha_x$ and so on for other angles:

$$\cos^2 \alpha_x + \cos^2 \alpha_y + \cos^2 \alpha_z = \frac{a_x^2}{a} + \frac{a_y^2}{a} + \frac{a_z^2}{a} = \frac{a_x^2 + a_y^2 + a_z^2}{a} = \frac{a}{a^2} = 1.$$

If two of a vector's direction angles (for e.g., $\alpha_x$ and $\alpha_y$) are less than $45{}^\circ$, then

$$\cos^2 \alpha_x + \cos^2 \alpha_y + \cos^2 \alpha_z > \cos^2 45{}^\circ + \cos^2 45{}^\circ + \cos^2 \alpha_z = \frac{1}{2} + \frac{1}{2} + \cos^2 \alpha_z \geq 1.$$

We obtained that $\cos^2 \alpha_x + \cos^2 \alpha_y + \cos^2 \alpha_z > 1$, and this inequality contradicts with the initial theorem.

**Answer**: the statement is correct.