Chapter 11 - Introduction to Three Dimensional Geometry

Let the given points be A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10).
Direction ratios of AB are $(-1-2, 2-3, -3-4) = (-3, -1, -7)$.
Direction ratios of BC are $(-4-(-1), 1-2, -10-(-3)) = (-3, -1, -7)$.
Since the direction ratios of AB and BC are proportional (in fact, identical), AB is parallel to BC. Since B is a common point, the points A, B, and C are collinear.
To find the ratio in which C divides AB, let the ratio be $k:1$.
Using the section formula for the x-coordinate of C:
$-4 = \frac{k(-1) + 1(2)}{k+1} \Rightarrow -4k - 4 = -k + 2 \Rightarrow -3k = 6 \Rightarrow k = -2$.
Since $k$ is negative, the division is external. The ratio is $2:1$ externally.

Let the vertices of the triangle be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.
Let the midpoints be $D(1, 5, -1)$, $E(0, 4, -2)$, and $F(2, 3, 4)$.
We have:
$x_1 + x_2 = 2(1) = 2$
$x_2 + x_3 = 2(0) = 0$
$x_3 + x_1 = 2(2) = 4$
Summing these: $2(x_1 + x_2 + x_3) = 6 \Rightarrow x_1 + x_2 + x_3 = 3$.
$x_3 = 3 - 2 = 1$, $x_1 = 3 - 0 = 3$, $x_2 = 3 - 4 = -1$.
Similarly for y-coordinates:
$y_1 + y_2 = 10$, $y_2 + y_3 = 8$, $y_3 + y_1 = 6$.
$2(y_1 + y_2 + y_3) = 24 \Rightarrow y_1 + y_2 + y_3 = 12$.
$y_3 = 12 - 10 = 2$, $y_1 = 12 - 8 = 4$, $y_2 = 12 - 6 = 6$.
Similarly for z-coordinates:
$z_1 + z_2 = -2$, $z_2 + z_3 = -4$, $z_3 + z_1 = 8$.
$2(z_1 + z_2 + z_3) = 2 \Rightarrow z_1 + z_2 + z_3 = 1$.
$z_3 = 1 - (-2) = 3$, $z_1 = 1 - (-4) = 5$, $z_2 = 1 - 8 = -7$.
The vertices are $A(3, 4, 5)$, $B(-1, 6, -7)$, and $C(1, 2, 3)$.
The centroid $G = (\frac{3-1+1}{3}, \frac{4+6+2}{3}, \frac{5-7+3}{3}) = (1, 4, \frac{1}{3})$.

Let the points be $A(0, -1, -7)$, $B(2, 1, -9)$, and $C(6, 5, -13)$.
Direction ratios of AB: $(2-0, 1-(-1), -9-(-7)) = (2, 2, -2)$.
Direction ratios of AC: $(6-0, 5-(-1), -13-(-7)) = (6, 6, -6)$.
Since the direction ratios are proportional ($6 = 3 \times 2$, etc.), the points are collinear.
To find the ratio in which A divides BC, let the ratio be $k:1$.
Using the x-coordinate of A:
$0 = \frac{k(6) + 1(2)}{k+1} \Rightarrow 6k + 2 = 0 \Rightarrow k = -\frac{1}{3}$.
The negative sign indicates external division. The ratio is $1:3$ externally.

One vertex is at the origin $O(0, 0, 0)$.
Since the edges coincide with the positive axes and have length 2, the adjacent vertices are $A(2, 0, 0)$, $B(0, 2, 0)$, and $C(0, 0, 2)$.
The vertices on the coordinate planes are $D(2, 2, 0)$, $E(0, 2, 2)$, and $F(2, 0, 2)$.
The vertex opposite to the origin is $G(2, 2, 2)$.
The coordinates are $(0,0,0), (2,0,0), (0,2,0), (0,0,2), (2,2,0), (0,2,2), (2,0,2), (2,2,2)$.

(a)-(iii): In the xy-plane, z is always 0.
(b)-(i): All coordinates are positive, so 1st octant.
(c)-(ii): x=0 defines the yz-plane.
(d)-(vi): For a line parallel to x-axis, y and z are constant.
(e)-(iv): Intersection of x=0 and y=0 is the z-axis.
(f)-(v): z=c is a plane at height c, parallel to xy-plane.
(g)-(viii): Intersection of x=a and y=b is a line parallel to z-axis.
(h)-(vii): Coordinates correspond to distances along axes.
(i)-(x): A ball is the interior of a sphere.
(j)-(ix): A disc is the interior of a circle.