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 \( |\vec{w}| \).

We are given that the vector is \(\vec{v}=(x, y)\) with magnitude \[ |\vec{v}|=\sqrt{x^2+y^2}=4, \] which implies \[ x^2+y^2=16. \] **a. For the vector \(\vec{u}=(-2x,-2y)\):** The magnitude of \(\vec{u}\) is \[ |\vec{u}|=\sqrt{(-2x)^2+(-2y)^2}=\sqrt{4x^2+4y^2}. \] This can be factored as \[ |\vec{u}|=\sqrt{4(x^2+y^2)}=2\sqrt{x^2+y^2}. \] Substitute \(x^2+y^2=16\): \[ |\vec{u}|=2\sqrt{16}=2\times4=8. \] **b. For the vector \(\vec{w}=(2x,2y)\):** The magnitude of \(\vec{w}\) is \[ |\vec{w}|=\sqrt{(2x)^2+(2y)^2}=\sqrt{4x^2+4y^2}. \] Similarly, this simplifies to \[ |\vec{w}|=\sqrt{4(x^2+y^2)}=2\sqrt{x^2+y^2}. \] Substitute \(x^2+y^2=16\): \[ |\vec{w}|=2\sqrt{16}=2\times4=8. \] Thus, the magnitudes are: \[ |\vec{u}|=8 \quad \text{and} \quad |\vec{w}|=8. \]