Two vectors are given as a = i + 2 j - k and b = 3 i - j + 2 k . A...

Given the vector equation $2\vec{a} + \vec{c} = \vec{b}$, we can solve for $\vec{c}$:
$\vec{c} = \vec{b} - 2\vec{a}$

Substituting the given vectors $\vec{a}$ and $\vec{b}$:
$\vec{c} = (3\hat{i} - \hat{j} + 2\hat{k}) - 2(\hat{i} + 2\hat{j} - \hat{k})$
$\Rightarrow \vec{c} = (3 - 2)\hat{i} + (-1 - 4)\hat{j} + (2 - (-2))\hat{k}$
$\Rightarrow \vec{c} = \hat{i} - 5\hat{j} + 4\hat{k}$

The magnitude of vector $\vec{c}$ is:
$|\vec{c}| = \sqrt{1^2 + (-5)^2 + 4^2} = \sqrt{1 + 25 + 16} = \sqrt{42}$

The cosine of the angle $\gamma$ made by vector $\vec{c}$ with the positive z-axis is given by the ratio of its z-component to its magnitude:
$\cos\gamma = \dfrac{c_z}{|\vec{c}|} = \dfrac{4}{\sqrt{42}}$

Answer: $\dfrac{4}{\sqrt{42}}$