The acute angle between the lines $\vec{r} = (4\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{j} - 3\hat{k})$ and $\frac{x-1}{1} = \frac{y+1}{-3} = \frac{z-2}{2}$ is

The acute angle between two lines is found using their direction vectors.

For Line 1: $\vec{r} = (4\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{j} - 3\hat{k})$

Direction vector: $\vec{b_1} = 2\hat{i} + \hat{j} - 3\hat{k}$

For Line 2: $\frac{x-1}{1} = \frac{y+1}{-3} = \frac{z-2}{2}$

Direction vector: $\vec{b_2} = 1\hat{i} - 3\hat{j} + 2\hat{k}$

---

The acute angle $\theta$ between two lines is given by:

$\cos\theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}||\vec{b_2}|}$

---

Calculate the dot product:

$\vec{b_1} \cdot \vec{b_2} = (2)(1) + (1)(-3) + (-3)(2)$

$= 2 - 3 - 6$

$= -7$

---

Calculate the magnitude of $\vec{b_1}$:

$|\vec{b_1}| = \sqrt{(2)^2 + (1)^2 + (-3)^2}$

$= \sqrt{4 + 1 + 9}$

$= \sqrt{14}$

Calculate the magnitude of $\vec{b_2}$:

$|\vec{b_2}| = \sqrt{(1)^2 + (-3)^2 + (2)^2}$

$= \sqrt{1 + 9 + 4}$

$= \sqrt{14}$

---

Substitute into the angle formula:

$\cos\theta = \frac{|-7|}{\sqrt{14} \times \sqrt{14}}$

$= \frac{7}{14}$

$= \frac{1}{2}$

---

When $\cos\theta = \frac{1}{2}$, the angle is $\theta = 60°$ or $\frac{\pi}{3}$ radians.

Therefore, the acute angle between the lines is $\frac{\pi}{3}$.