Introduction to Three Dimensional Geometry

Given: A point lies on the x-axis.

Any point on the x-axis is of the form $(x, 0, 0)$.

Therefore, the y-coordinate and z-coordinate of the point are both $\mathbf{0}$.

Given: A point lies in the XZ-plane.

Any point in the XZ-plane has its y-coordinate equal to zero, because the XZ-plane is defined as the set of all points where $y = 0$.

Therefore, the y-coordinate of the point is $\mathbf{0}$.

The sign convention for octants is:

| Octant | x | y | z |
|--------|---|---|---|
| I | + | + | + |
| II | − | + | + |
| III | − | − | + |
| IV | + | − | + |
| V | + | + | − |
| VI | − | + | − |
| VII | − | − | − |
| VIII | + | − | − |

1. $(1, 2, 3)$: x>0,\ y>0,\ z>0 → Octant I
2. $(4, -2, 3)$: x>0,\ y<0,\ z>0 → Octant IV
3. $(4, -2, -5)$: x>0,\ y<0,\ z<0 → Octant VIII
4. $(4, 2, -5)$: x>0,\ y>0,\ z<0 → Octant V
5. $(-4, 2, -5)$: x<0,\ y>0,\ z<0 → Octant VI
6. $(-4, 2, 5)$: x<0,\ y>0,\ z>0 → Octant II
7. $(-3, -1, 6)$: x<0,\ y<0,\ z>0 → Octant III
8. $(-2, -4, -7)$: x<0,\ y<0,\ z<0 → Octant VII

(i) The x-axis and y-axis taken together determine a plane known as the XY-plane (also called the xy-plane).

(ii) The coordinates of points in the XY-plane are of the form $\mathbf{(x,\ y,\ 0)}$, since the z-coordinate is zero for every point in the XY-plane.

(iii) Coordinate planes divide the space into eight octants.

Formula used: Distance between $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ is
$$PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

(i) $P(2,3,5)$ and $Q(4,3,1)$:
$$PQ = \sqrt{(4-2)^2+(3-3)^2+(1-5)^2} = \sqrt{4+0+16} = \sqrt{20} = 2\sqrt{5}$$

(ii) $P(-3,7,2)$ and $Q(2,4,-1)$:
$$PQ = \sqrt{(2-(-3))^2+(4-7)^2+(-1-2)^2} = \sqrt{25+9+9} = \sqrt{43}$$

(iii) $P(-1,3,-4)$ and $Q(1,-3,4)$:
$$PQ = \sqrt{(1-(-1))^2+(-3-3)^2+(4-(-4))^2} = \sqrt{4+36+64} = \sqrt{104} = 2\sqrt{26}$$

(iv) $P(2,-1,3)$ and $Q(-2,1,3)$:
$$PQ = \sqrt{(-2-2)^2+(1-(-1))^2+(3-3)^2} = \sqrt{16+4+0} = \sqrt{20} = 2\sqrt{5}$$

Let $A(-2,3,5)$, $B(1,2,3)$, $C(7,0,-1)$.

Using the distance formula:
$$AB = \sqrt{(1+2)^2+(2-3)^2+(3-5)^2} = \sqrt{9+1+4} = \sqrt{14}$$

$$BC = \sqrt{(7-1)^2+(0-2)^2+(-1-3)^2} = \sqrt{36+4+16} = \sqrt{56} = 2\sqrt{14}$$

$$AC = \sqrt{(7+2)^2+(0-3)^2+(-1-5)^2} = \sqrt{81+9+36} = \sqrt{126} = 3\sqrt{14}$$

Now, $AB + BC = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = AC$.

Since $AB + BC = AC$, the three points are collinear.

(i) Let $A(0,7,-10)$, $B(1,6,-6)$, $C(4,9,-6)$.

$$AB = \sqrt{(1-0)^2+(6-7)^2+(-6+10)^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$$

$$BC = \sqrt{(4-1)^2+(9-6)^2+(-6+6)^2} = \sqrt{9+9+0} = \sqrt{18} = 3\sqrt{2}$$

$$CA = \sqrt{(0-4)^2+(7-9)^2+(-10+6)^2} = \sqrt{16+4+16} = \sqrt{36} = 6$$

Since $AB = BC = 3\sqrt{2}$ and $CA = 6$, two sides are equal.

Hence, the triangle is isosceles. ✓

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(ii) Let $A(0,7,10)$, $B(-1,6,6)$, $C(-4,9,6)$.

$$AB^2 = (-1-0)^2+(6-7)^2+(6-10)^2 = 1+1+16 = 18$$

$$BC^2 = (-4+1)^2+(9-6)^2+(6-6)^2 = 9+9+0 = 18$$

$$CA^2 = (0+4)^2+(7-9)^2+(10-6)^2 = 16+4+16 = 36$$

Check: $AB^2 + BC^2 = 18 + 18 = 36 = CA^2$.

Since $AB^2 + BC^2 = CA^2$, by the converse of Pythagoras' theorem, the triangle is a right angled triangle (right angle at B). ✓

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(iii) Let $A(-1,2,1)$, $B(1,-2,5)$, $C(4,-7,8)$, $D(2,-3,4)$.

For a parallelogram, opposite sides must be equal: $AB = DC$ and $BC = AD$.

$$AB = \sqrt{(1+1)^2+(-2-2)^2+(5-1)^2} = \sqrt{4+16+16} = \sqrt{36} = 6$$

$$DC = \sqrt{(4-2)^2+(-7+3)^2+(8-4)^2} = \sqrt{4+16+16} = \sqrt{36} = 6$$

$$BC = \sqrt{(4-1)^2+(-7+2)^2+(8-5)^2} = \sqrt{9+25+9} = \sqrt{43}$$

$$AD = \sqrt{(2+1)^2+(-3-2)^2+(4-1)^2} = \sqrt{9+25+9} = \sqrt{43}$$

Since $AB = DC$ and $BC = AD$, the quadrilateral is a parallelogram. ✓

Let $P(x,y,z)$ be any point equidistant from $A(1,2,3)$ and $B(3,2,-1)$.

Given condition: $PA = PB$, i.e., $PA^2 = PB^2$.

$$PA^2 = (x-1)^2+(y-2)^2+(z-3)^2$$

$$PB^2 = (x-3)^2+(y-2)^2+(z+1)^2$$

Setting $PA^2 = PB^2$:
$$(x-1)^2+(y-2)^2+(z-3)^2 = (x-3)^2+(y-2)^2+(z+1)^2$$

$$(x-1)^2+(z-3)^2 = (x-3)^2+(z+1)^2$$

$$x^2-2x+1+z^2-6z+9 = x^2-6x+9+z^2+2z+1$$

$$-2x-6z+10 = -6x+2z+10$$

$$4x - 8z = 0$$

$$\boxed{x - 2z = 0}$$

This is the required equation of the set of points.

Let $P(x,y,z)$ be any point such that $PA + PB = 10$.

$$PA = \sqrt{(x-4)^2+y^2+z^2}, \quad PB = \sqrt{(x+4)^2+y^2+z^2}$$

Given: $PA + PB = 10$, so $PA = 10 - PB$.

Squaring both sides:
$$PA^2 = 100 - 20\cdot PB + PB^2$$

$$(x-4)^2+y^2+z^2 = 100 - 20\cdot PB + (x+4)^2+y^2+z^2$$

$$x^2-8x+16 = 100 - 20\cdot PB + x^2+8x+16$$

$$-8x = 100 - 20\cdot PB + 8x$$

$$20\cdot PB = 100 + 16x$$

$$PB = \frac{100+16x}{20} = 5 + \frac{4x}{5}$$

Squaring again:
$$PB^2 = \left(5+\frac{4x}{5}\right)^2$$

$$(x+4)^2+y^2+z^2 = 25 + 8x + \frac{16x^2}{25}$$

$$x^2+8x+16+y^2+z^2 = 25+8x+\frac{16x^2}{25}$$

$$x^2+16+y^2+z^2 = 25+\frac{16x^2}{25}$$

Multiplying throughout by 25:
$$25x^2+400+25y^2+25z^2 = 625+16x^2$$

$$9x^2+25y^2+25z^2 = 225$$

Dividing by 225:
$$\boxed{\frac{x^2}{25}+\frac{y^2}{9}+\frac{z^2}{9} = 1}$$

This is the required equation of the set of points P.

Given: $A(3,-1,2)$, $B(1,2,-4)$, $C(-1,1,2)$. Let $D(x,y,z)$ be the fourth vertex.

Concept: In a parallelogram, the diagonals bisect each other. So the midpoint of diagonal $AC$ = midpoint of diagonal $BD$.

Midpoint of $AC$:
$$\left(\frac{3+(-1)}{2},\ \frac{-1+1}{2},\ \frac{2+2}{2}\right) = (1,\ 0,\ 2)$$

Midpoint of $BD$:
$$\left(\frac{1+x}{2},\ \frac{2+y}{2},\ \frac{-4+z}{2}\right)$$

Setting them equal:
$$\frac{1+x}{2} = 1 \Rightarrow x = 1$$
$$\frac{2+y}{2} = 0 \Rightarrow y = -2$$
$$\frac{-4+z}{2} = 2 \Rightarrow z = 8$$

The coordinates of the fourth vertex $D$ are $\mathbf{(1,\ -2,\ 8)}$.

Given: $A(0,0,6)$, $B(0,4,0)$, $C(6,0,0)$.

Let $D$, $E$, $F$ be the midpoints of $BC$, $CA$, $AB$ respectively.

Midpoint D of BC:
$$D = \left(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\right) = (3,2,0)$$

Median AD:
$$AD = \sqrt{(3-0)^2+(2-0)^2+(0-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7$$

Midpoint E of CA:
$$E = \left(\frac{6+0}{2},\frac{0+0}{2},\frac{0+6}{2}\right) = (3,0,3)$$

Median BE:
$$BE = \sqrt{(3-0)^2+(0-4)^2+(3-0)^2} = \sqrt{9+16+9} = \sqrt{34}$$

Midpoint F of AB:
$$F = \left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right) = (0,2,3)$$

Median CF:
$$CF = \sqrt{(0-6)^2+(2-0)^2+(3-0)^2} = \sqrt{36+4+9} = \sqrt{49} = 7$$

The lengths of the medians are $AD = 7$, $BE = \sqrt{34}$, $CF = 7$.

Given: Origin $O(0,0,0)$ is the centroid of $\triangle PQR$ with $P(2a,2,6)$, $Q(-4,3b,-10)$, $R(8,14,2c)$.

Formula: Centroid $= \left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3},\ \dfrac{z_1+z_2+z_3}{3}\right)$

Setting each coordinate of the centroid equal to 0:

x-coordinate:
$$\frac{2a+(-4)+8}{3} = 0 \Rightarrow 2a+4 = 0 \Rightarrow a = -2$$

y-coordinate:
$$\frac{2+3b+14}{3} = 0 \Rightarrow 3b+16 = 0 \Rightarrow b = -\frac{16}{3}$$

z-coordinate:
$$\frac{6+(-10)+2c}{3} = 0 \Rightarrow 2c-4 = 0 \Rightarrow c = 2$$

Therefore, $a = -2$, $b = -\dfrac{16}{3}$, $c = 2$.

Let $P(x,y,z)$ be any point satisfying the given condition.

$$PA^2 = (x-3)^2+(y-4)^2+(z-5)^2$$
$$= x^2-6x+9+y^2-8y+16+z^2-10z+25$$

$$PB^2 = (x+1)^2+(y-3)^2+(z+7)^2$$
$$= x^2+2x+1+y^2-6y+9+z^2+14z+49$$

Adding:
$$PA^2+PB^2 = 2x^2+2y^2+2z^2-4x-14y+4z+109$$

Given $PA^2+PB^2 = k^2$:
$$2x^2+2y^2+2z^2-4x-14y+4z+109 = k^2$$

$$\boxed{2x^2+2y^2+2z^2-4x-14y+4z = k^2-109}$$

This is the required equation of the set of points P.