Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>

Solution:

Given, two vectors are

u = 2i - 4j

v = 3i - 8j

We have to find the angle between the given vectors.

The dot product is given by u.v

= (2i - 4j).(3i - 8j)

= 2(3) + (-4)(-8)

= 6 + 32

= 38

The magnitude of the vectors is given by

|u| = \(\sqrt{(2)^{2}+(-4)^{2}}\\=\sqrt{4+16}\\=\sqrt{20}\)

|u| = 2√5

|v| = \(\sqrt{(3)^{2}+(-8)^{2}}\\=\sqrt{9+64}\\=\sqrt{73}\)

The angle between the vectors is given by

\(\theta =cos^{-1}\frac{\vec{u}\vec{v}}{\left | u \right |\left | v \right |}\)

\(\theta =cos^{-1}\frac{38}{(2\sqrt{5})(\sqrt{73})}\)

\(\theta =cos^{-1}\frac{19}{(\sqrt{365})}\)

\(\theta =cos^{-1}\frac{19}{19.1049}\)

\(\theta =cos^{-1}(0.9945)\)

\(\theta =6^{\circ}\)

Therefore, the angle between the vectors is \(\theta =6^{\circ}\)

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Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>

Summary:

The angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8> is \(\theta =6^{\circ}\).