Chapter 11 - Introduction to Three Dimensional Geometry

(i) $x>0, y<0, z>0$ corresponds to Octant IV.
(ii) $x<0, y>0, z>0$ corresponds to Octant II.
(iii) $x<0, y<0, z<0$ corresponds to Octant VII.
(iv) $x<0, y>0, z<0$ corresponds to Octant VI.

Signs of coordinates determine the octant:
(i) $(+,+,+)$: Octant I
(ii) $(+,-,+)$: Octant IV
(iii) $(+,-,-)$: Octant VIII
(iv) $(+,+,-)$: Octant V
(v) $(-,+,+)$: Octant II
(vi) $(-,-,+)$: Octant III
(vii) $(+,-,-)$: Octant VIII
(viii) $(-,+,-)$: Octant VI

The foot of the perpendicular from $P(x, y, z)$ on the x-axis is $(x, 0, 0)$, on the y-axis is $(0, y, 0)$, and on the z-axis is $(0, 0, z)$.
(i) For $P(3, 4, 2)$: $A(3, 0, 0), B(0, 4, 0), C(0, 0, 2)$.
(ii) For $P(-5, 3, 7)$: $A(-5, 0, 0), B(0, 3, 0), C(0, 0, 7)$.
(iii) For $P(4, -3, -5)$: $A(4, 0, 0), B(0, -3, 0), C(0, 0, -5)$.

The foot of the perpendicular from $P(x, y, z)$ on the xy-plane is $(x, y, 0)$, on the yz-plane is $(0, y, z)$, and on the zx-plane is $(x, 0, z)$.
(i) For $P(3, 4, 5)$: $A(3, 4, 0), B(0, 4, 5), C(3, 0, 5)$.
(ii) For $P(-5, 3, 7)$: $A(-5, 3, 0), B(0, 3, 7), C(-5, 0, 7)$.
(iii) For $P(4, -3, -5)$: $A(4, -3, 0), B(0, -3, -5), C(4, 0, -5)$.

Distance $= \sqrt{(-3-2)^2 + (0-0)^2 + (0-0)^2} = \sqrt{(-5)^2} = 5$ units.

Distance $= \sqrt{6^2 + 6^2 + 7^2} = \sqrt{36 + 36 + 49} = \sqrt{121} = 11$ units.

Let the point be $P(x, y, z)$ where $z = \sqrt{1 - x^2 - y^2}$.
Distance from origin $OP = \sqrt{x^2 + y^2 + z^2}$.
$OP = \sqrt{x^2 + y^2 + (\sqrt{1 - x^2 - y^2})^2} = \sqrt{x^2 + y^2 + 1 - x^2 - y^2} = \sqrt{1} = 1$ unit.

$AB = \sqrt{(2-1)^2 + (-4+1)^2 + (5-3)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$.
$BC = \sqrt{(5-2)^2 + (-13+4)^2 + (11-5)^2} = \sqrt{9 + 81 + 36} = \sqrt{126} = 3\sqrt{14}$.
$AC = \sqrt{(5-1)^2 + (-13+1)^2 + (11-3)^2} = \sqrt{16 + 144 + 64} = \sqrt{224} = 4\sqrt{14}$.
Since $AB + BC = \sqrt{14} + 3\sqrt{14} = 4\sqrt{14} = AC$, the points are collinear.

Let the fourth vertex be $D(x, y, z)$. The midpoints of diagonals AC and BD coincide.
Midpoint of AC: $(\frac{6-2}{2}, \frac{-2+2}{2}, \frac{4+4}{2}) = (2, 0, 4)$.
Midpoint of BD: $(\frac{2+x}{2}, \frac{4+y}{2}, \frac{-8+z}{2})$.
Equating coordinates:
$\frac{2+x}{2} = 2 \Rightarrow x = 2$.
$\frac{4+y}{2} = 0 \Rightarrow y = -4$.
$\frac{-8+z}{2} = 4 \Rightarrow z = 16$.
So, $D(2, -4, 16)$.

$AB^2 = (2-0)^2 + (3-4)^2 + (-1-1)^2 = 4 + 1 + 4 = 9$.
$BC^2 = (4-2)^2 + (5-3)^2 + (0+1)^2 = 4 + 4 + 1 = 9$.
$AC^2 = (4-0)^2 + (5-4)^2 + (0-1)^2 = 16 + 1 + 1 = 18$.
Since $AB^2 + BC^2 = 9 + 9 = 18 = AC^2$, the triangle is right-angled at B.

Let the third vertex be $(x, y, z)$. Centroid is $(0, 0, 0)$.
$\frac{2 + 0 + x}{3} = 0 \Rightarrow x = -2$.
$\frac{4 - 2 + y}{3} = 0 \Rightarrow y = -2$.
$\frac{6 - 5 + z}{3} = 0 \Rightarrow z = -1$.
Third vertex is $(-2, -2, -1)$.

The centroid of the triangle formed by the midpoints of the sides is the same as the centroid of the original triangle.
Centroid $G = (\frac{1+3-1}{3}, \frac{2+0+1}{3}, \frac{-3+1-4}{3}) = (\frac{3}{3}, \frac{3}{3}, \frac{-6}{3}) = (1, 1, -2)$.

Let vertices be $A(x_1, y_1, z_1), B(x_2, y_2, z_2), C(x_3, y_3, z_3)$.
Midpoints: $D(5, 7, 11), E(0, 8, 5), F(2, 3, -1)$.
$x_1+x_2=10, x_2+x_3=0, x_3+x_1=4$. Sum: $2(x_1+x_2+x_3)=14 \Rightarrow \sum x = 7$.
$x_3 = 7-10 = -3, x_1 = 7-0 = 7, x_2 = 7-4 = 3$.
Similarly for y: $\sum y = 18$. $y_3 = 18-14=4, y_1 = 18-16=2, y_2 = 18-6=12$.
For z: $\sum z = 15$. $z_3 = 15-22=-7, z_1 = 15-10=5, z_2 = 15-(-2)=17$.
Vertices are $(7, 2, 5), (3, 12, 17), (-3, 4, -7)$.

Midpoint of AC = Midpoint of BD.
Midpoint of AC: $(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}) = (1.5, 2.5, 2.5)$.
Let $D(x, y, z)$. Midpoint of BD: $(\frac{-1+x}{2}, \frac{-2+y}{2}, \frac{-1+z}{2})$.
$-1+x = 3 \Rightarrow x = 4$.
$-2+y = 5 \Rightarrow y = 7$.
$-1+z = 5 \Rightarrow z = 6$.
Vertex D is $(4, 7, 6)$.

Let P and Q be the points of trisection. P divides AB in 1:2, Q divides AB in 2:1.
P: $(\frac{1(5)+2(2)}{3}, \frac{1(-8)+2(1)}{3}, \frac{1(3)+2(-3)}{3}) = (\frac{9}{3}, \frac{-6}{3}, \frac{-3}{3}) = (3, -2, -1)$.
Q: $(\frac{2(5)+1(2)}{3}, \frac{2(-8)+1(1)}{3}, \frac{2(3)+1(-3)}{3}) = (\frac{12}{3}, \frac{-15}{3}, \frac{3}{3}) = (4, -5, 1)$.

Centroid is $(0, 0, 0)$.
$\frac{a - 2 + 4}{3} = 0 \Rightarrow a + 2 = 0 \Rightarrow a = -2$.
$\frac{1 + b + 7}{3} = 0 \Rightarrow b + 8 = 0 \Rightarrow b = -8$.
$\frac{3 - 5 + c}{3} = 0 \Rightarrow c - 2 = 0 \Rightarrow c = 2$.

Calculate lengths of sides AB and AC.
$AB = \sqrt{(5-2)^2 + (6-2)^2 + (9+3)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.
$AC = \sqrt{(2-2)^2 + (7-2)^2 + (9+3)^2} = \sqrt{0 + 25 + 144} = \sqrt{169} = 13$.
Since $AB = AC$, the triangle is isosceles. The internal bisector of angle A is the median to the base BC.
Therefore, D is the midpoint of BC.
$D = (\frac{5+2}{2}, \frac{6+7}{2}, \frac{9+9}{2}) = (\frac{7}{2}, \frac{13}{2}, 9)$.