Given vectors a = 3i − 2j + 4k and b = 2i + j + 3k, find the magnitude and direction cosines for a - b and 3a - 2b.

$$\vec{a} - \vec{b} = (3\vec{i} - 2\vec{j} + 4\vec{k}) - (2\vec{i} + \vec{j} + 3\vec{k})$$ $$\vec{a} - \vec{b} = (3-2)\vec{i} + (-2-1)\vec{j} + (4-3)\vec{k}$$ $$\vec{a} - \vec{b} = \vec{i} - 3\vec{j} + \vec{k}$$

$$|\vec{a} - \vec{b}| = \sqrt{(1)^2 + (-3)^2 + (1)^2}$$ $$|\vec{a} - \vec{b}| = \sqrt{1 + 9 + 1}$$ $$|\vec{a} - \vec{b}| = \sqrt{11}$$

The direction cosines are $\frac{x}{|\vec{r}|}$, $\frac{y}{|\vec{r}|}$, $\frac{z}{|\vec{r}|}$.

$$\cos\alpha = \frac{1}{\sqrt{11}}$$ $$\cos\beta = \frac{-3}{\sqrt{11}}$$ $$\cos\gamma = \frac{1}{\sqrt{11}}$$

The magnitude is $\boxed{\sqrt{11}}$ and the direction cosines are $\boxed{\left(\frac{1}{\sqrt{11}}, \frac{-3}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right)}$.

$$3\vec{a} - 2\vec{b} = (9\vec{i} - 6\vec{j} + 12\vec{k}) - (4\vec{i} + 2\vec{j} + 6\vec{k})$$ $$3\vec{a} - 2\vec{b} = (9-4)\vec{i} + (-6-2)\vec{j} + (12-6)\vec{k}$$ $$3\vec{a} - 2\vec{b} = 5\vec{i} - 8\vec{j} + 6\vec{k}$$

$$|3\vec{a} - 2\vec{b}| = \sqrt{(5)^2 + (-8)^2 + (6)^2}$$ $$|3\vec{a} - 2\vec{b}| = \sqrt{25 + 64 + 36}$$ $$|3\vec{a} - 2\vec{b}| = \sqrt{125} = \sqrt{25 \times 5}$$ $$|3\vec{a} - 2\vec{b}| = 5\sqrt{5}$$

$$\cos\alpha = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}$$ $$\cos\beta = \frac{-8}{5\sqrt{5}}$$ $$\cos\gamma = \frac{6}{5\sqrt{5}}$$

The magnitude is $\boxed{5\sqrt{5}}$ and the direction cosines are $\boxed{\left(\frac{1}{\sqrt{5}}, \frac{-8}{5\sqrt{5}}, \frac{6}{5\sqrt{5}}\right)}$.