We are tasked with finding the acute angle between lines $AM$ and $BN$, where $ABCD$ is a parallelogram with specific dimensions and points defined as $M$ and $N$. Here's the step-by-step process to solve this:

1. Given Information:

• $AB = 3\sqrt{2}$
• $AD = 8$
• $\angle BAD = 45^\circ$
• $CM : MD = 1 : 2$, so $M$ divides $CD$ in the ratio $1:2$.
• $AN : ND = 3 : 1$, so $N$ divides $AD$ in the ratio $3:1$.

We need to find the acute angle between vectors $\overrightarrow{AM}$ and $\overrightarrow{BN}$.

2. Coordinate Setup:

We can place the parallelogram on the coordinate plane. Let:

• $A = (0, 0)$
• $B = (3\sqrt{2}, 0)$
• $D = (0, 8)$

Now, using the fact that $\angle BAD = 45^\circ$, we can determine the coordinates of point $C$, which is directly opposite to $A$ in the parallelogram:

• The vector $\overrightarrow{AB}$ is along the x-axis, and the vector $\overrightarrow{AD}$ makes a $45^\circ$ angle with $AB$. Therefore, $C = (3\sqrt{2}, 8)$.

3. Finding Coordinates of $M$ and $N$:

• Point $M$ on $CD$: Since $M$ divides $CD$ in the ratio $1:2$, use the section formula to find its coordinates: $M = \left( \frac{1(0) + 2(3\sqrt{2})}{1+2}, \frac{1(8) + 2(8)}{1+2} \right) = \left( 2\sqrt{2}, 8 \right)$

• Point $N$ on $AD$: Since $N$ divides $AD$ in the ratio $3:1$, its coordinates are: $N = \left( \frac{3(0) + 1(0)}{3+1}, \frac{3(8) + 1(0)}{3+1} \right) = (0, 6)$

4. Vectors $\overrightarrow{AM}$ and $\overrightarrow{BN}$:

Now, we calculate the vectors $\overrightarrow{AM}$ and $\overrightarrow{BN}$:

• $\overrightarrow{AM} = (2\sqrt{2}, 8) - (0, 0) = (2\sqrt{2}, 8)$
• $\overrightarrow{BN} = (0, 6) - (3\sqrt{2}, 0) = (-3\sqrt{2}, 6)$

5. Dot Product to Find the Angle:

The dot product formula between two vectors $\mathbf{u} = (x_1, y_1)$ and $\mathbf{v} = (x_2, y_2)$ is: $\mathbf{u} \cdot \mathbf{v} = x_1 x_2 + y_1 y_2$ Applying this to $\overrightarrow{AM} = (2\sqrt{2}, 8)$ and $\overrightarrow{BN} = (-3\sqrt{2}, 6)$: $\overrightarrow{AM} \cdot \overrightarrow{BN} = (2\sqrt{2})(-3\sqrt{2}) + (8)(6) = -12 + 48 = 36$

Next, compute the magnitudes of the vectors: $|\overrightarrow{AM}| = \sqrt{(2\sqrt{2})^2 + 8^2} = \sqrt{8 + 64} = \sqrt{72} = 6\sqrt{2}$ $|\overrightarrow{BN}| = \sqrt{(-3\sqrt{2})^2 + 6^2} = \sqrt{18 + 36} = \sqrt{54} = 3\sqrt{6}$

Finally, use the dot product formula to find the cosine of the angle $\theta$ between the vectors: $\cos \theta = \frac{\overrightarrow{AM} \cdot \overrightarrow{BN}}{|\overrightarrow{AM}| |\overrightarrow{BN}|} = \frac{36}{(6\sqrt{2})(3\sqrt{6})} = \frac{36}{18\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

Thus, the angle $\theta$ is: $\theta = \cos^{-1} \left( \frac{2\sqrt{3}}{3} \right) = 30^\circ$

Final Answer:

The acute angle between $AM$ and $BN$ is $30^\circ$.

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Would you like more details or have any questions?

Here are five related questions to deepen understanding:

1. How do you calculate the angle between two vectors in 3D?
2. How does the section formula apply to other shapes?
3. What is the significance of the dot product in finding angles?
4. How can the concept of vector projection help in geometry problems?
5. What happens if the ratio in which a point divides a line changes?

Tip: The dot product is zero when two vectors are perpendicular to each other, giving an easy way to check for right angles!