State and prove the Pythagorean theorem of two vectors and verify this for u = (1, -1) and v = (1, 1).

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Pythagorem theorem of two vectors states that if two vectors say \(\vec{u}\) and \(\vec{v}\) are orthogonal (i.e. their dot product us zero \(\vec{u} . \vec{v} = 0\) then

\(||\vec{u} + \vec{v}||^2 = ||\vec{u}||^2 + ||\vec{v}||^2\)

Proof:

\(||\vec{u} + \vec{v}||^2\)

\(= (\vec{u} + \vec{v}) (\vec{u} + \vec{v}) \)

\(= \vec{u}.\vec{u} + \vec{v}.\vec{v} + 2\vec{u}.\vec{v}\)

\(= \vec{u}.\vec{u} + \vec{v}.\vec{v} + 0\)

\(= ||\vec{u}||^2 + ||\vec{v}||^2\)

          proved

Given

\(\vec{u}\) = (1, -1) and \(\vec{v}\) = (1, 1)

\(\vec{u} + \vec{v}\) = (1 + 1, -1 + 1) = (2, 0)

\(||\vec{u} + \vec{v}||^2\) = \((\sqrt{(2)^2 + 0})^2\) = 4

Also,

\(||\vec{u}||^2\) = \((\sqrt{(1)^2 + (-1)^2})^2\) = 2

\(||\vec{v}||^2\) = \((\sqrt{(1)^2 + (1)^2})^2\) = 2

\(||\vec{u}||^2 + ||\vec{v}||^2\) = 2 + 2 = 4 = \(||\vec{u} + \vec{v}||^2\)

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