Three Dimensional Geometry
Miscellaneous Exercise Solutions
Key Concepts:
Midpoint: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})$
Centroid: $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$
Midpoint: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})$
Centroid: $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$
Q1
1. Three vertices of parallelogram ABCD are A(3, -1, 2), B(1, 2, -4), C(-1, 1, 2). Find vertex D.
Property
Diagonals of a parallelogram bisect each other. Midpoint of AC = Midpoint of BD.
Midpoint AC
$(\frac{3+(-1)}{2}, \frac{-1+1}{2}, \frac{2+2}{2}) = (\frac{2}{2}, \frac{0}{2}, \frac{4}{2}) = (1, 0, 2)$.
Midpoint BD
Let D be $(x, y, z)$. Midpoint is $(\frac{1+x}{2}, \frac{2+y}{2}, \frac{-4+z}{2})$.
Equate
$\frac{1+x}{2} = 1 \Rightarrow x=1$
$\frac{2+y}{2} = 0 \Rightarrow y=-2$
$\frac{-4+z}{2} = 2 \Rightarrow z=8$
$\frac{2+y}{2} = 0 \Rightarrow y=-2$
$\frac{-4+z}{2} = 2 \Rightarrow z=8$
Coordinates of D: $(1, -2, 8)$
Q2
2. Find lengths of medians of triangle with vertices A(0, 0, 6), B(0, 4, 0), C(6, 0, 0).
Median AD
D is mid of BC: $(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}) = (3, 2, 0)$.
Length $AD = \sqrt{(3-0)^2 + (2-0)^2 + (0-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7$.
Length $AD = \sqrt{(3-0)^2 + (2-0)^2 + (0-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7$.
Median BE
E is mid of AC: $(\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}) = (3, 0, 3)$.
Length $BE = \sqrt{(3-0)^2 + (0-4)^2 + (3-0)^2} = \sqrt{9+16+9} = \sqrt{34}$.
Length $BE = \sqrt{(3-0)^2 + (0-4)^2 + (3-0)^2} = \sqrt{9+16+9} = \sqrt{34}$.
Median CF
F is mid of AB: $(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}) = (0, 2, 3)$.
Length $CF = \sqrt{(0-6)^2 + (2-0)^2 + (3-0)^2} = \sqrt{36+4+9} = \sqrt{49} = 7$.
Length $CF = \sqrt{(0-6)^2 + (2-0)^2 + (3-0)^2} = \sqrt{36+4+9} = \sqrt{49} = 7$.
Lengths: $7, \sqrt{34}, 7$
Q3
3. Origin is centroid of $\Delta PQR$ with vertices P(2a, 2, 6), Q(-4, 3b, -10), R(8, 14, 2c). Find a, b, c.
Centroid
Formula: $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}) = (0, 0, 0)$.
x-coord
$\frac{2a – 4 + 8}{3} = 0 \Rightarrow 2a + 4 = 0 \Rightarrow a = -2$.
y-coord
$\frac{2 + 3b + 14}{3} = 0 \Rightarrow 3b + 16 = 0 \Rightarrow b = -16/3$.
z-coord
$\frac{6 – 10 + 2c}{3} = 0 \Rightarrow 2c – 4 = 0 \Rightarrow c = 2$.
$a = -2, \quad b = -16/3, \quad c = 2$
Q4
4. Find equation of set of points P such that $PA^2 + PB^2 = k^2$, where A(3, 4, 5) and B(-1, 3, -7).
Let P
Let $P(x, y, z)$.
PA²
$(x-3)^2 + (y-4)^2 + (z-5)^2 = x^2 – 6x + 9 + y^2 – 8y + 16 + z^2 – 10z + 25$.
PB²
$(x+1)^2 + (y-3)^2 + (z+7)^2 = x^2 + 2x + 1 + y^2 – 6y + 9 + z^2 + 14z + 49$.
Sum
$2x^2 + 2y^2 + 2z^2 – 4x – 14y + 4z + 109 = k^2$.
$2x^2 + 2y^2 + 2z^2 – 4x – 14y + 4z + (109 – k^2) = 0$