The velocity of a particle is given as \(\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}\) m/s. The magnitude of acceleration at point \((1, 2, 4)\) is _______ m/s\(^2\).

The velocity vector is given by \(\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} = -x \hat{i} + 2y \hat{j} - z \hat{k}\).

The components of acceleration are:

\(a_x = v_x \dfrac{\partial v_x}{\partial x} + v_y \dfrac{\partial v_x}{\partial y} + v_z \dfrac{\partial v_x}{\partial z} = (-x)(-1) + 0 + 0 = x\)

\(a_y = v_x \dfrac{\partial v_y}{\partial x} + v_y \dfrac{\partial v_y}{\partial y} + v_z \dfrac{\partial v_y}{\partial z} = 0 + (2y)(2) + 0 = 4y\)

\(a_z = v_x \dfrac{\partial v_z}{\partial x} + v_y \dfrac{\partial v_z}{\partial y} + v_z \dfrac{\partial v_z}{\partial z} = 0 + 0 + (-z)(-1) = z\)

The acceleration vector is \(\vec{a} = x \hat{i} + 4y \hat{j} + z \hat{k}\).

At the point \((1, 2, 4)\), substituting \(x = 1\), \(y = 2\), and \(z = 4\):

\(\vec{a} = 1 \hat{i} + 4(2) \hat{j} + 4 \hat{k} = \hat{i} + 8 \hat{j} + 4 \hat{k}\)

The magnitude of acceleration is:

\(|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9\) m/s\(^2\)

Answer: \(9\)