Chapter 11: Introduction to Three Dimensional Geometry

From Fig 11.3, $$F$$ is the foot of the perpendicular dropped from $$P$$ to the $$XZ$$–plane (the plane $$y = 0$$).

Projecting a point onto the $$XZ$$–plane keeps its $$x$$– and $$z$$–coordinates and sets the $$y$$–coordinate to zero, so

$$x_F = x_P = 2, \; y_F = 0, \; z_F = z_P = 5.$$

Therefore, the coordinates of $$F$$ are $$F(2, 0, 5)$$.

Each octant of three-dimensional space is identified by the signs of the three coordinates $$(x, y, z)$$ of any point lying in it. The eight octants correspond to the sign-combinations below.

For the point $$(-3, 1, 2)$$: here $$x = -3 \lt 0$$, $$y = 1 \gt 0$$ and $$z = 2 \gt 0$$, so the sign pattern is $$(-,\, +,\, +)$$. From the table this is octant II.

For the point $$(-3, 1, -2)$$: here $$x = -3 \lt 0$$, $$y = 1 \gt 0$$ and $$z = -2 \lt 0$$, so the sign pattern is $$(-,\, +,\, -)$$. From the table this is octant VI.

The distance between two points $$(x_1, y_1, z_1)$$ and $$(x_2, y_2, z_2)$$ in space is given by the distance formula

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Here $$P(1, -3, 4)$$ and $$Q(-4, 1, 2)$$, so

$$PQ = \sqrt{(-4 - 1)^2 + (1 - (-3))^2 + (2 - 4)^2}$$

$$PQ = \sqrt{(-5)^2 + (4)^2 + (-2)^2} = \sqrt{25 + 16 + 4}$$

$$PQ = \sqrt{45} = 3\sqrt{5}$$ units.

Three points are collinear (lie on one straight line) if the sum of the lengths of two of the segments joining them equals the length of the third (the longest) segment.

Using the distance formula for each pair:

$$PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$

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

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

Now add the two shorter lengths:

$$PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR$$

Since $$PQ + QR = PR$$, the points $$P$$, $$Q$$ and $$R$$ are collinear.

A triangle is right-angled if (and only if) the square of its longest side equals the sum of the squares of the other two sides (Pythagoras). So compute the squares of the three side lengths.

$$AB^2 = (10 - 3)^2 + (20 - 6)^2 + (30 - 9)^2 = 7^2 + 14^2 + 21^2 = 49 + 196 + 441 = 686$$

$$BC^2 = (25 - 10)^2 + (-41 - 20)^2 + (5 - 30)^2 = 15^2 + (-61)^2 + (-25)^2 = 225 + 3721 + 625 = 4571$$

$$CA^2 = (3 - 25)^2 + (6 - (-41))^2 + (9 - 5)^2 = (-22)^2 + 47^2 + 4^2 = 484 + 2209 + 16 = 2709$$

Test the Pythagoras relation for every possible choice of hypotenuse:

$$AB^2 + CA^2 = 686 + 2709 = 3395 \neq 4571 = BC^2$$

$$AB^2 + BC^2 = 686 + 4571 = 5257 \neq 2709 = CA^2$$

$$BC^2 + CA^2 = 4571 + 2709 = 7280 \neq 686 = AB^2$$

None of the three combinations satisfies the relation, so no angle of the triangle is a right angle.

Let $$P(x, y, z)$$ be any point of the required set, with $$A(3, 4, 5)$$ and $$B(-1, 3, -7)$$.

By the distance formula,

$$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2$$

$$PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2$$

Add them, grouping the $$x$$, $$y$$ and $$z$$ terms separately:

$$(x - 3)^2 + (x + 1)^2 = (x^2 - 6x + 9) + (x^2 + 2x + 1) = 2x^2 - 4x + 10$$

$$(y - 4)^2 + (y - 3)^2 = (y^2 - 8y + 16) + (y^2 - 6y + 9) = 2y^2 - 14y + 25$$

$$(z - 5)^2 + (z + 7)^2 = (z^2 - 10z + 25) + (z^2 + 14z + 49) = 2z^2 + 4z + 74$$

Therefore

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

The given condition $$PA^2 + PB^2 = 2k^2$$ gives

$$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = 2k^2$$

$$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = 2k^2 - 109$$

This is the required equation of the set of points; it represents a sphere.

The $$x$$-axis is the line common to the XY-plane and the XZ-plane — every point on the $$x$$-axis lies in both of these planes.

A point lying in the XY-plane has its $$z$$-coordinate equal to $$0$$, and a point lying in the XZ-plane has its $$y$$-coordinate equal to $$0$$.

Since a point on the $$x$$-axis lies in both planes simultaneously, it must satisfy both conditions. Hence its $$y$$-coordinate and $$z$$-coordinate are both zero; such a point is of the form $$(x, 0, 0)$$.

The XZ-plane is the plane containing the $$x$$-axis and the $$z$$-axis. It consists of exactly those points whose displacement measured along the $$y$$-direction is zero.

Therefore every point lying in the XZ-plane has its $$y$$-coordinate equal to zero, and the point is of the form $$(x, 0, z)$$.

The octant in which a point lies is fixed by the signs of its coordinates $$(x, y, z)$$. The standard correspondence is: I $$(+,+,+)$$, II $$(-,+,+)$$, III $$(-,-,+)$$, IV $$(+,-,+)$$, V $$(+,+,-)$$, VI $$(-,+,-)$$, VII $$(-,-,-)$$, VIII $$(+,-,-)$$.

Reading the sign pattern of each given point:

Label the points $$A(-2, 3, 5)$$, $$B(1, 2, 3)$$ and $$C(7, 0, -1)$$. Three points are collinear if the sum of the lengths of two of the segments joining them equals the length of the third (the longest) segment.

Apply the distance formula to each pair:

$$AB = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$

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

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

Now

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

Since $$AB + BC = AC$$, the three points $$A$$, $$B$$ and $$C$$ are collinear.

Let $$P(x, y, z)$$ be a point of the required set, and let $$A(1, 2, 3)$$ and $$B(3, 2, -1)$$. “Equidistant” means $$PA = PB$$, so $$PA^2 = PB^2$$.

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

Expand both sides:

$$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 + 2z + 1$$

Cancel the terms common to both sides ($$x^2$$, $$y^2$$, $$z^2$$, $$-4y$$ and $$+4$$):

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

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

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

Collect like terms:

$$6x - 2x = 2z + 6z \quad\Rightarrow\quad 4x = 8z$$

$$x - 2z = 0$$

This is the required equation of the set of points; it represents a plane.

Let $$P(x, y, z)$$ be a point of the set, with $$A(4, 0, 0)$$ and $$B(-4, 0, 0)$$. The given condition is $$PA + PB = 10$$, i.e.

$$\sqrt{(x - 4)^2 + y^2 + z^2} + \sqrt{(x + 4)^2 + y^2 + z^2} = 10$$

Move one radical to the right side:

$$\sqrt{(x - 4)^2 + y^2 + z^2} = 10 - \sqrt{(x + 4)^2 + y^2 + z^2}$$

Square both sides:

$$(x - 4)^2 + y^2 + z^2 = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2} + (x + 4)^2 + y^2 + z^2$$

Since $$(x - 4)^2 - (x + 4)^2 = -16x$$, cancelling $$y^2$$ and $$z^2$$ gives

$$-16x = 100 - 20\sqrt{(x + 4)^2 + y^2 + z^2}$$

$$20\sqrt{(x + 4)^2 + y^2 + z^2} = 100 + 16x$$

Divide by $$4$$:

$$5\sqrt{(x + 4)^2 + y^2 + z^2} = 25 + 4x$$

Square both sides again:

$$25\left[(x + 4)^2 + y^2 + z^2\right] = (25 + 4x)^2$$

$$25x^2 + 200x + 400 + 25y^2 + 25z^2 = 625 + 200x + 16x^2$$

Bring all terms to one side and simplify:

$$25x^2 - 16x^2 + 25y^2 + 25z^2 + 400 - 625 = 0$$

$$9x^2 + 25y^2 + 25z^2 - 225 = 0$$

This is the required equation of the set of points.

To prove $$ABCD$$ is a parallelogram, show that both pairs of opposite sides are equal. To prove it is not a rectangle, show that its diagonals are unequal (a rectangle must have equal diagonals).

Side lengths:

$$AB = \sqrt{(-1 - 1)^2 + (-2 - 2)^2 + (-1 - 3)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

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

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

$$DA = \sqrt{(1 - 4)^2 + (2 - 7)^2 + (3 - 6)^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$$

Since $$AB = CD = 6$$ and $$BC = DA = \sqrt{43}$$, both pairs of opposite sides are equal, so $$ABCD$$ is a parallelogram.

Diagonals:

$$AC = \sqrt{(2 - 1)^2 + (3 - 2)^2 + (2 - 3)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$$

$$BD = \sqrt{(4 - (-1))^2 + (7 - (-2))^2 + (6 - (-1))^2} = \sqrt{25 + 81 + 49} = \sqrt{155}$$

Since $$AC = \sqrt{3} \neq \sqrt{155} = BD$$, the diagonals are unequal. Therefore $$ABCD$$ is a parallelogram but not a rectangle.

Let $$P(x, y, z)$$ be a point of the required set. Its distances from $$A(3, 4, -5)$$ and $$B(-2, 1, 4)$$ are equal, so $$PA = PB$$, hence $$PA^2 = PB^2$$.

$$(x - 3)^2 + (y - 4)^2 + (z + 5)^2 = (x + 2)^2 + (y - 1)^2 + (z - 4)^2$$

Expand both sides:

$$x^2 - 6x + 9 + y^2 - 8y + 16 + z^2 + 10z + 25 = x^2 + 4x + 4 + y^2 - 2y + 1 + z^2 - 8z + 16$$

Cancel $$x^2$$, $$y^2$$ and $$z^2$$ from both sides, and collect the constants:

$$-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$$

Bring every term to one side:

$$0 = (4x + 6x) + (-2y + 8y) + (-8z - 10z) + (21 - 50)$$

$$10x + 6y - 18z - 29 = 0$$

This is the required equation of the set of points; it represents a plane (the perpendicular bisector plane of segment $$AB$$).

Let the coordinates of $$C$$ be $$(x, y, z)$$. The centroid $$G$$ of a triangle with vertices $$A$$, $$B$$ and $$C$$ is the average of their coordinates:

$$G = \left(\frac{x_A + x_B + x_C}{3},\ \frac{y_A + y_B + y_C}{3},\ \frac{z_A + z_B + z_C}{3}\right)$$

Here $$A(3, -5, 7)$$, $$B(-1, 7, -6)$$ and $$G(1, 1, 1)$$. Equating each coordinate of the centroid to $$1$$:

$$\frac{3 + (-1) + x}{3} = 1 \;\Rightarrow\; 2 + x = 3 \;\Rightarrow\; x = 1$$

$$\frac{-5 + 7 + y}{3} = 1 \;\Rightarrow\; 2 + y = 3 \;\Rightarrow\; y = 1$$

$$\frac{7 + (-6) + z}{3} = 1 \;\Rightarrow\; 1 + z = 3 \;\Rightarrow\; z = 2$$

Hence the coordinates of $$C$$ are $$(1, 1, 2)$$.

Let the fourth vertex be $$D(x, y, z)$$. In a parallelogram the diagonals bisect each other, so the midpoint of diagonal $$AC$$ must coincide with the midpoint of diagonal $$BD$$.

Midpoint of $$AC$$, with $$A(3, -1, 2)$$ and $$C(-1, 1, 2)$$:

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

Midpoint of $$BD$$, with $$B(1, 2, -4)$$ and $$D(x, y, z)$$:

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

Equate the two midpoints coordinate by coordinate:

$$\frac{1 + x}{2} = 1 \;\Rightarrow\; 1 + x = 2 \;\Rightarrow\; x = 1$$

$$\frac{2 + y}{2} = 0 \;\Rightarrow\; 2 + y = 0 \;\Rightarrow\; y = -2$$

$$\frac{-4 + z}{2} = 2 \;\Rightarrow\; -4 + z = 4 \;\Rightarrow\; z = 8$$

Hence the fourth vertex is $$D(1, -2, 8)$$.

Let the vertices be $$A(0, 0, 6)$$, $$B(0, 4, 0)$$ and $$C(6, 0, 0)$$. A median joins a vertex to the midpoint of the opposite side, so first find the midpoints, then use the distance formula.

Median $$AD$$ (from $$A$$ to the midpoint $$D$$ of $$BC$$):

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

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

Median $$BE$$ (from $$B$$ to the midpoint $$E$$ of $$AC$$):

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

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

Median $$CF$$ (from $$C$$ to the midpoint $$F$$ of $$AB$$):

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

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

Hence the lengths of the three medians are $$7$$, $$\sqrt{34}$$ and $$7$$ units.

The centroid of triangle $$PQR$$ is the average of the coordinates of its vertices:

$$\left(\frac{x_P + x_Q + x_R}{3},\ \frac{y_P + y_Q + y_R}{3},\ \frac{z_P + z_Q + z_R}{3}\right)$$

It is given that this centroid is the origin $$(0, 0, 0)$$, so each coordinate of the centroid equals $$0$$.

From the $$x$$-coordinate:

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

From the $$y$$-coordinate:

$$\frac{2 + 3b + 14}{3} = 0 \;\Rightarrow\; 3b + 16 = 0 \;\Rightarrow\; b = -\frac{16}{3}$$

From the $$z$$-coordinate:

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

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

Let $$P(x, y, z)$$ be any point of the required set, with $$A(3, 4, 5)$$ and $$B(-1, 3, -7)$$.

By the distance formula,

$$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2$$

$$PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2$$

Add them, grouping the $$x$$, $$y$$ and $$z$$ terms separately:

$$(x - 3)^2 + (x + 1)^2 = (x^2 - 6x + 9) + (x^2 + 2x + 1) = 2x^2 - 4x + 10$$

$$(y - 4)^2 + (y - 3)^2 = (y^2 - 8y + 16) + (y^2 - 6y + 9) = 2y^2 - 14y + 25$$

$$(z - 5)^2 + (z + 7)^2 = (z^2 - 10z + 25) + (z^2 + 14z + 49) = 2z^2 + 4z + 74$$

So

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

The given condition $$PA^2 + PB^2 = k^2$$ yields

$$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = k^2$$

$$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z - k^2 + 109 = 0$$

This is the required equation of the set of points; it represents a sphere.