Question

(a) Prove that for the two vectors \( \vec{a} \) and \( \vec{b} \), \( |\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}| \):

Hint:

For proving inequalities involving vectors, use the properties of the dot product and trigonometric bounds.

Answer 1

The dot product of two vectors is given by: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta, \] where \( \theta \) is the angle between the vectors. The absolute value of \( \cos \theta \) is always less than or equal to 1, i.e., \( |\cos \theta| \leq 1 \). Therefore: \[ |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta| \leq |\vec{a}| |\vec{b}|. \]