Subject: Applied Mathematics 4

Topic: Vector Spaces

Difficulty: Medium

$ ||u|| = \sqrt{(-4)^2 + 2^2 + 1^2} = \sqrt{16+4+1} = \sqrt{21} \\ ||v|| = \sqrt{8^2 + (-4)^2 + (-2)^2} = \sqrt{64+16+4} = \sqrt{84} \\ ||u||.||v|| = \sqrt{21}\sqrt{84} = 42 $

Here, $\bar{u} = -4 \bar{i} + 2 \bar{j} + \bar{k} \\ \bar{v} = 8 \bar{i} - 4 \bar{j} + (-2) \bar{k} $

$ \therefore \bar{u}.\bar{v} = (-4)8 + 2(-4) + 1(-2) = -42 \\ \therefore |\bar{u}.\bar{v}| = 42 \\ \therefore |\bar{u}.\bar{v}| = ||u||.||v|| $

Hence, Cauchy-Schwarz inequality is verified $[ |\bar{u}.\bar{v}| \leq ||u||.||v|| ]$ .