The problem involves a parallelogram ABCD with three given vertices A(-3, 2, x), B(3, -3, 1), and C(5, x, 2), and asks for the following:

a) Determine the fourth vertex D. b) Find the value of $x$ such that the length of vector $\overrightarrow{AD}$ equals $\sqrt{5}$. c) Calculate the angle between the diagonals of the parallelogram if $x = -3$.

Let me break down the steps to solve these:

Part a) Finding vertex D:

For a parallelogram, the midpoint of the diagonals coincide. Thus, the midpoint of diagonal $AC$ must be the same as the midpoint of diagonal $BD$.

• The midpoint of diagonal $AC$ can be calculated as the average of the coordinates of A and C.
• Similarly, the midpoint of diagonal $BD$ can be calculated as the average of the coordinates of B and D.

We will set these midpoints equal and solve for $D$.

Part b) Finding $x$ so that $\overrightarrow{AD} = \sqrt{5}$:

The vector $\overrightarrow{AD}$ is the difference between the coordinates of $A$ and $D$. We need to set the length of this vector equal to $\sqrt{5}$, and solve for $x$.

Part c) Calculating the angle between diagonals:

The angle between two vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$ can be found using the dot product formula: $\cos(\theta) = \frac{\overrightarrow{AC} \cdot \overrightarrow{BD}}{|\overrightarrow{AC}| |\overrightarrow{BD}|}$ Here we calculate the vectors, take their dot product, and use the magnitude to find the angle.

Let me start solving this step by step.

Part a) Finding D:

We know the coordinates of A(-3, 2, x) and C(5, x, 2). The midpoint of AC is:

$M_{AC} = \left(\frac{-3 + 5}{2}, \frac{2 + x}{2}, \frac{x + 2}{2}\right) = \left(1, \frac{2 + x}{2}, \frac{x + 2}{2}\right)$

Similarly, the midpoint of BD is:

$M_{BD} = \left(\frac{3 + x_1}{2}, \frac{-3 + y_1}{2}, \frac{1 + z_1}{2}\right)$

Let me proceed to solve these to determine D. sympy import symbols, Eq, solve

Define symbols for x1, y1, z1 (coordinates of D)

x1, y1, z1 = symbols('x1 y1 z1')

Coordinates of A and C

A = (-3, 2, 'x') # A = (-3, 2, x) C = (5, 'x', 2) # C = (5, x, 2)

Midpoint of AC = M_AC = ( (A_x + C_x) / 2, (A_y + C_y) / 2, (A_z + C_z) / 2 )

M_AC = ((A[0] + C[0]) / 2, (A[1] + C[1]) / 2, (A[2] + C[2]) / 2)

Coordinates of B

B = (3, -3, 1) # B = (3, -3, 1)

Midpoint of BD = M_BD = ((B_x + D_x) / 2, (B_y + D_y) / 2, (B_z + D_z) / 2 )

M_BD = ((B[0] + x1) / 2, (B[1] + y1) / 2, (B[2] + z1) / 2)

Set midpoint of AC equal to midpoint of BD

eq1 = Eq(M_AC[0], M_BD[0]) eq2 = Eq(M_AC[1], M_BD[1]) eq3 = Eq(M_AC[2], M_BD[2])

Solve for x1, y1, z1 (coordinates of D)

solution = solve((eq1, eq2, eq3), (x1, y1, z1)) solutionIt seems that there was an error because of the presence of the variable $x$ in the coordinates of A and C. I will correct the approach and ensure that the variable ( x \