We are given three points $A(0,0,-1)$, $B(1,2,1)$, and $C(-2,-1,1)$ in a 3D space with an orthonormal coordinate system. a. Find the coordinates of the vectors $\vec{AB}$, $\vec{AC}$, and $\vec{BC}$. b. Show that $\triangle ABC$ is a right-angled isosceles triangle. c. Find the coordinates of point $D$ such that the quadrilateral $ABDC$ is a square. We are also given the equation $9(y^2 - 10y) + 16(x^2 - 6x) + 225 = 0$. a. Show that this equation represents an ellipse. b. Find the coordinates of the center, vertices, and foci of this ellipse.

1. Problem Description

We are given three points $A(0,0,-1)$, $B(1,2,1)$, and $C(-2,-1,1)$ in a 3D space with an orthonormal coordinate system.

a. Find the coordinates of the vectors $\vec{AB}$, $\vec{AC}$, and $\vec{BC}$.

b. Show that $\triangle ABC$ is a right-angled isosceles triangle.

c. Find the coordinates of point $D$ such that the quadrilateral $ABDC$ is a square.

We are also given the equation $9(y^2 - 10y) + 16(x^2 - 6x) + 225 = 0$.

a. Show that this equation represents an ellipse.

b. Find the coordinates of the center, vertices, and foci of this ellipse.

2. Solution Steps

Part 1: Triangle in 3D space

a. Finding the vectors:

$\vec{AB} = B - A = (1, 2, 1) - (0, 0, -1) = (1, 2, 2)$

$\vec{AC} = C - A = (-2, -1, 1) - (0, 0, -1) = (-2, -1, 2)$

$\vec{BC} = C - B = (-2, -1, 1) - (1, 2, 1) = (-3, -3, 0)$

b. Showing that $\triangle ABC$ is a right-angled isosceles triangle:

First, we check for orthogonality using the dot product:

$\vec{AB} \cdot \vec{AC} = (1)(-2) + (2)(-1) + (2)(2) = -2 - 2 + 4 = 0$

Since $\vec{AB} \cdot \vec{AC} = 0$, $\vec{AB}$ and $\vec{AC}$ are orthogonal, so $\angle BAC = 90^{\circ}$.

Next, we find the lengths of $\vec{AB}$ and $\vec{AC}$:

$||\vec{AB}|| = \sqrt{(1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

$||\vec{AC}|| = \sqrt{(-2)^2 + (-1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Since $||\vec{AB}|| = ||\vec{AC}|| = 3$, the triangle is isosceles.

Therefore, $\triangle ABC$ is a right-angled isosceles triangle.

c. Finding the point D:

Since $ABDC$ is a square, $\vec{AD} = \vec{BC}$.

So, $D = A + \vec{BC} = (0, 0, -1) + (-3, -3, 0) = (-3, -3, -1)$.

Part 2: Ellipse

$9(y^2 - 10y) + 16(x^2 - 6x) + 225 = 0$

$9(y^2 - 10y + 25) + 16(x^2 - 6x + 9) + 225 = 9(25) + 16(9)$

$9(y-5)^2 + 16(x-3)^2 + 225 = 225 + 144$

$9(y-5)^2 + 16(x-3)^2 = 144$

$\frac{(y-5)^2}{16} + \frac{(x-3)^2}{9} = 1$

$\frac{(x-3)^2}{9} + \frac{(y-5)^2}{16} = 1$

a. Showing that the equation represents an ellipse:

The equation is in the form $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$, where $a^2 = 16$ and $b^2 = 9$, and $a > b$. Thus, the equation represents an ellipse with center $(h, k)$.

b. Finding the center, vertices, and foci:

Center: $(h, k) = (3, 5)$

$a^2 = 16$, so $a = 4$.

$b^2 = 9$, so $b = 3$.

$c^2 = a^2 - b^2 = 16 - 9 = 7$, so $c = \sqrt{7}$.

Since $a$ is associated with $y$, the major axis is vertical.

Vertices: $(h, k \pm a) = (3, 5 \pm 4) = (3, 9)$ and $(3, 1)$.

Foci: $(h, k \pm c) = (3, 5 \pm \sqrt{7}) = (3, 5 + \sqrt{7})$ and $(3, 5 - \sqrt{7})$.

3. Final Answer

Part 1:

a. $\vec{AB} = (1, 2, 2)$, $\vec{AC} = (-2, -1, 2)$, $\vec{BC} = (-3, -3, 0)$

b. $\triangle ABC$ is a right-angled isosceles triangle.

c. $D = (-3, -3, -1)$

Part 2:

a. The equation represents an ellipse.

b. Center: $(3, 5)$, Vertices: $(3, 9)$ and $(3, 1)$, Foci: $(3, 5 + \sqrt{7})$ and $(3, 5 - \sqrt{7})$.