ABCD is a parallelogram. AB = 3√2, AD = 8, ∠BAD = 45◦, M is a point on CD, CM : MD = 1 : 2. N is a point on AD, AN : ND = 3 : 1. Find:
a) the length of AM; b) the length of BN;
c) the acute angle between AM and BN.

We are given a parallelogram \( ABCD \), where:

- \( AB = 3\sqrt{2} \)
- \( AD = 8 \)
- \( \angle BAD = 45^\circ \)
- \( M \) is a point on \( CD \) such that \( CM:MD = 1:2 \)
- \( N \) is a point on \( AD \) such that \( AN:ND = 3:1 \)

We need to find:

a) The length of \( AM \)  
b) The length of \( BN \)  
c) The acute angle between \( AM \) and \( BN \)

### Step 1: Set up a coordinate system
Let’s place point \( A \) at the origin \( (0, 0) \), point \( B \) along the \( x \)-axis, and point \( D \) along the \( y \)-axis. From the given information:

- \( A(0, 0) \)
- \( B(3\sqrt{2}, 0) \)
- \( D(0, 8) \)

Using the property of a parallelogram, \( C \), the opposite vertex of \( A \), can be found by adding the vectors \( \overrightarrow{AD} \) and \( \overrightarrow{AB} \):

\[
C = B + D = (3\sqrt{2}, 8)
\]

### Step 2: Find coordinates of points M and N

- **For point \( M \)** on \( CD \):  
  The ratio \( CM:MD = 1:2 \) implies that \( M \) divides the line segment \( CD \) in the ratio 1:2. Using the section formula, the coordinates of \( M \) are:

  \[
  M = \left( \frac{2(0) + 1(3\sqrt{2})}{3}, \frac{2(8) + 1(0)}{3} \right) = \left( \frac{3\sqrt{2}}{3}, \frac{16}{3} \right) = \left( \sqrt{2}, \frac{16}{3} \right)
  \]

- **For point \( N \)** on \( AD \):  
  The ratio \( AN:ND = 3:1 \) implies that \( N \) divides the line segment \( AD \) in the ratio 3:1. Using the section formula, the coordinates of \( N \) are:

  \[
  N = \left( \frac{1(0) + 3(0)}{4}, \frac{1(8) + 3(0)}{4} \right) = (0, 2)
  \]

### Step 3: Calculate the length of \( AM \)

The length of \( AM \) can be calculated using the distance formula:

\[
AM = \sqrt{(x_M - x_A)^2 + (y_M - y_A)^2} = \sqrt{(\sqrt{2} - 0)^2 + \left( \frac{16}{3} - 0 \right)^2}
\]

\[
AM = \sqrt{2 + \left( \frac{16}{3} \right)^2} = \sqrt{2 + \frac{256}{9}} = \sqrt{\frac{18}{9} + \frac{256}{9}} = \sqrt{\frac{274}{9}} = \frac{\sqrt{274}}{3}
\]

### Step 4: Calculate the length of \( BN \)

The length of \( BN \) can be calculated using the distance formula:

\[
BN = \sqrt{(x_B - x_N)^2 + (y_B - y_N)^2} = \sqrt{(3\sqrt{2} - 0)^2 + (0 - 2)^2}
\]

\[
BN = \sqrt{(3\sqrt{2})^2 + (-2)^2} = \sqrt{18 + 4} = \sqrt{22}
\]

### Step 5: Calculate the acute angle between \( AM \) and \( BN \)

To find the acute angle \( \theta \) between \( AM \) and \( BN \), we use the formula for the dot product:

\[
\cos \theta = \frac{\vec{AM} \cdot \vec{BN}}{|\vec{AM}| |\vec{BN}|}
\]

First, compute the vectors \( \vec{AM} \) and \( \vec{BN} \):

- \( \vec{AM} = (\sqrt{2} - 0, \frac{16}{3} - 0) = (\sqrt{2}, \frac{16}{3}) \)
- \( \vec{BN} = (0 - 3\sqrt{2}, 2 - 0) = (-3\sqrt{2}, 2) \)

The dot product \( \vec{AM} \cdot \vec{BN} \) is:

\[
\vec{AM} \cdot \vec{BN} = (\sqrt{2})(-3\sqrt{2}) + \left(\frac{16}{3}\right)(2) = -6 + \frac{32}{3} = \frac{-18

ABCD is a parallelogram. AB = 3√2, AD = 8, ∠BAD = 45◦, M is a point on CD, CM : MD = 1 : 2. N is a point on AD, AN : ND = 3 : 1. Find: a) the length of AM; b) the length of BN; c) the acute angle between AM and BN.

Solve a parallelogram geometry problem involving vectors and coordinate geometry. Learn how to find the length of AM and BN, and calculate the acute angle between vectors AM and BN using the distance and dot product formulas. Ideal for students in Grades 10-12.

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