19. Given vector \( \vec{v}=(x, y) \) with \( |\vec{v}|=4 \). If a. vector \( \vec{u}=(-2 x,-2 y) \), so find \( |\vec{u}| \). b. vector \( \vec{w}=(2 x, 2 y) \), so find \( |\vec{w}| \).

To find the magnitudes of vectors \( \vec{u} \) and \( \vec{w} \), we start with the information given about \( \vec{v} \). Since the magnitude of \( \vec{v} = (x, y) \) is \( |\vec{v}| = 4 \), we know: \[ \sqrt{x^2 + y^2} = 4 \] Squaring both sides gives: \[ x^2 + y^2 = 16 \] a. For vector \( \vec{u} = (-2x, -2y) \): \[ |\vec{u}| = \sqrt{(-2x)^2 + (-2y)^2} = \sqrt{4x^2 + 4y^2} = \sqrt{4(x^2 + y^2)} = \sqrt{4 \cdot 16} = \sqrt{64} = 8 \] So, \( |\vec{u}| = 8 \). b. For vector \( \vec{w} = (2x, 2y) \): \[ |\vec{w}| = \sqrt{(2x)^2 + (2y)^2} = \sqrt{4x^2 + 4y^2} = \sqrt{4(x^2 + y^2)} = \sqrt{4 \cdot 16} = \sqrt{64} = 8 \] Thus, \( |\vec{w}| = 8 \).