证明由 $\mathbf{r} = (5i + 2j + 4k) + \lambda(i + 3j + k)$ 与 $\mathbf{r} = (3i + j + k) + \mu(4i + 7j + 5k)$ 给出的两条直线相交，并求出它们的交点。 计算这两条直线之间的锐角。

1. Set up a system of linear equations by equating the $x$, $y$, and $z$ components of the two line equations.

$5 + \lambda = 3 + 4\mu \implies \lambda - 4\mu = -2 \quad \text{--- (1)}$
$2 + 3\lambda = 1 + 7\mu \implies 3\lambda - 7\mu = -1 \quad \text{--- (2)}$
$4 + \lambda = 1 + 5\mu \implies \lambda - 5\mu = -3 \quad \text{--- (3)}$

2. Solve any two of the equations simultaneously to find the values of $\lambda$ and $\mu$.

From (1), $\lambda = 4\mu - 2$.
Substitute into (2):
$3(4\mu - 2) - 7\mu = -1$
$12\mu - 6 - 7\mu = -1$
$5\mu = 5 \implies \mu = 1$

Substitute $\mu = 1$ back to find $\lambda$:
$\lambda = 4(1) - 2 = 2$

3. Verify that these parameter values satisfy the third equation to confirm the lines intersect.

Substitute $\lambda = 2$ and $\mu = 1$ into the left hand side of (3):
$\text{LHS} = \lambda - 5\mu = 2 - 5(1) = -3$
$\text{RHS} = -3$
Since LHS = RHS, the system is consistent, which proves that the two lines intersect.

4. Substitute the parameter back into the respective line equation to find the point of intersection.

Substitute $\lambda = 2$ into the first line equation:
$\mathbf{r} = (5\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) + 2(\mathbf{i} + 3\mathbf{j} + \mathbf{k})$
$\mathbf{r} = 7\mathbf{i} + 8\mathbf{j} + 6\mathbf{k}$
(Alternatively, substitute $\mu = 1$ into the second line equation, which yields the same position vector.)

The point of intersection is $(7, 8, 6)$.

1. Identify the direction vectors of the two given lines.

Let the direction vectors of the lines be $\mathbf{d}_1$ and $\mathbf{d}_2$.

$\mathbf{d}_1 = \mathbf{i} + 3\mathbf{j} + \mathbf{k} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$
$\mathbf{d}_2 = 4\mathbf{i} + 7\mathbf{j} + 5\mathbf{k} = \begin{pmatrix} 4 \\ 7 \\ 5 \end{pmatrix}$

2. Calculate the scalar product and the magnitudes of the direction vectors.

$\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(4) + (3)(7) + (1)(5) = 4 + 21 + 5 = 30$

$|\mathbf{d}_1| = \sqrt{1^2 + 3^2 + 1^2} = \sqrt{11}$
$|\mathbf{d}_2| = \sqrt{4^2 + 7^2 + 5^2} = \sqrt{16 + 49 + 25} = \sqrt{90}$

3. Use the scalar product to find the acute angle between the lines.

Let $\theta$ be the acute angle between the lines.

$\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|}$
$\cos \theta = \frac{30}{\sqrt{11}\sqrt{90}} = \frac{30}{\sqrt{990}}$

$\theta = \cos^{-1}\left(\frac{30}{\sqrt{990}}\right) \approx 17.548^\circ \approx 17.5^\circ$ (to 1 d.p.)