Mathematics: Class 11 — Chapter 11: Introduction to Three Dimensional Geometry

Consider three mutually perpendicular planes intersecting at a point $O$ (the origin). These planes intersect along three mutually perpendicular lines $X'OX$, $Y'OY$, and $Z'OZ$, which are called the x-axis, y-axis, and z-axis respectively.

• Coordinate Planes: The planes $XOY$, $YOZ$, and $ZOX$ are called the XY-plane, YZ-plane, and ZX-plane respectively.
• Octants: The three coordinate planes divide the space into eight parts, known as octants, denoted by Roman numerals I to VIII.

Signs of Coordinates in Eight Octants:

Key Spatial Rules:

• The coordinates of the origin $O$ are $(0, 0, 0)$.
• Any point on the $x$-axis has coordinates of the form $(x, 0, 0)$.
• Any point on the $y$-axis has coordinates of the form $(0, y, 0)$.
• Any point on the $z$-axis has coordinates of the form $(0, 0, z)$.
• Any point in the $XY$-plane has coordinates of the form $(x, y, 0)$.
• Any point in the $YZ$-plane has coordinates of the form $(0, y, z)$.
• Any point in the $ZX$-plane has coordinates of the form $(x, 0, z)$.

To find the coordinates of any point $P$ in space:

1. Drop a perpendicular $PM$ from $P$ to the $XY$-plane, where $M$ is the foot of the perpendicular.
2. From $M$, draw a perpendicular $ML$ to the $x$-axis, meeting it at $L$.
3. Let $OL = x$, $LM = y$, and $MP = z$. The ordered triplet $(x, y, z)$ represents the coordinates of point $P$.

Alternatively, we can construct a rectangular parallelepiped through $P$ with faces parallel to the coordinate planes. The lengths of the edges along the $x$, $y$, and $z$ axes starting from the origin represent the coordinates.

Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be two points in space. Construct a rectangular parallelepiped with $PQ$ as its diagonal, such that its faces are parallel to the coordinate planes.

In the right-angled triangle $PAQ$ (where $\angle PAQ = 90^\circ$):

In the right-angled triangle $ANQ$ (where $\angle ANQ = 90^\circ$):

Substituting (2) into (1) yields:

Since the edges of the parallelepiped are parallel to the coordinate axes, their lengths are:

• $AN = |x_2 - x_1|$
• $PA = |y_2 - y_1|$
• $NQ = |z_2 - z_1|$

Thus, the distance $PQ$ is given by:

Distance from the Origin: If $P$ is the origin $O(0,0,0)$, the distance to any point $Q(x, y, z)$ is:

Mathematical Problem Statement: Find the distance between the points $P(1, -3, 4)$ and $Q(-4, 1, 2)$.

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Step-by-Step Algorithmic Solution Trace:

1. Identify coordinates: $(x_1, y_1, z_1) = (1, -3, 4)$ and $(x_2, y_2, z_2) = (-4, 1, 2)$.
2. Apply the 3D distance formula:$$PQ = \sqrt{(-4 - 1)^2 + (1 - (-3))^2 + (2 - 4)^2}$$
3. Simplify the terms inside the radical:$$PQ = \sqrt{(-5)^2 + (4)^2 + (-2)^2}$$$$PQ = \sqrt{25 + 16 + 4} = \sqrt{45}$$
4. Simplify the square root:$$PQ = 3\sqrt{5}$$

Final Evaluation Output: $3\sqrt{5} \text{ units}$

Mathematical Problem Statement: Show that the points $P(-2, 3, 5)$, $Q(1, 2, 3)$, and $R(7, 0, -1)$ are collinear.

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Step-by-Step Algorithmic Solution Trace:

1. Calculate distance $PQ$:$$PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$
2. Calculate distance $QR$:$$QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}$$
3. Calculate distance $PR$:$$PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$$
4. Check the collinearity condition ($PQ + QR = PR$):$$\sqrt{14} + 2\sqrt{14} = 3\sqrt{14}$$

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

Mathematical Problem Statement: Find the equation of the set of points $P(x, y, z)$ such that $PA^2 + PB^2 = 2k^2$, where $A(3, 4, 5)$ and $B(-1, 3, -7)$ are given points.

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Step-by-Step Algorithmic Solution Trace:

1. Express $PA^2$ using the squared distance formula:$$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - 5)^2$$
2. Express $PB^2$ using the squared distance formula:$$PB^2 = (x + 1)^2 + (y - 3)^2 + (z + 7)^2$$
3. Substitute these expressions into the given relation $PA^2 + PB^2 = 2k^2$:$$[(x - 3)^2 + (y - 4)^2 + (z - 5)^2] + [(x + 1)^2 + (y - 3)^2 + (z + 7)^2] = 2k^2$$
4. Expand each term:$$(x^2 - 6x + 9 + y^2 - 8y + 16 + z^2 - 10z + 25) + (x^2 + 2x + 1 + y^2 - 6y + 9 + z^2 + 14z + 49) = 2k^2$$
5. Combine like terms:$$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + 109 = 2k^2$$
6. Rearrange to standard form:$$2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = 2k^2 - 109$$

Final Evaluation Output: $2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = 2k^2 - 109$

Mathematical Problem Statement: Find the equation of the set of the points $P$ such that its distances from the points $A(3, 4, -5)$ and $B(-2, 1, 4)$ are equal.

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Step-by-Step Algorithmic Solution Trace:

1. Let the coordinates of point $P$ be $(x, y, z)$. The condition is $PA = PB$, which implies $PA^2 = PB^2$.
2. Write the squared distance from $P$ to $A(3, 4, -5)$:$$PA^2 = (x - 3)^2 + (y - 4)^2 + (z + 5)^2$$
3. Write the squared distance from $P$ to $B(-2, 1, 4)$:$$PB^2 = (x + 2)^2 + (y - 1)^2 + (z - 4)^2$$
4. Equate $PA^2$ and $PB^2$:$$(x - 3)^2 + (y - 4)^2 + (z + 5)^2 = (x + 2)^2 + (y - 1)^2 + (z - 4)^2$$
5. Expand both sides:$$(x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 10z + 25) = (x^2 + 4x + 4) + (y^2 - 2y + 1) + (z^2 - 8z + 16)$$
6. Cancel $x^2$, $y^2$, and $z^2$ from both sides and simplify:$$-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$$
7. Group all terms on one side:$$10x + 6y - 18z - 29 = 0$$

Final Evaluation Output: $10x + 6y - 18z - 29 = 0$

Mathematical Problem Statement: The centroid of a triangle $ABC$ is at the point $(1, 1, 1)$. If the coordinates of $A$ and $B$ are $(3, -5, 7)$ and $(-1, 7, -6)$, respectively, find the coordinates of the point $C$.

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Step-by-Step Algorithmic Solution Trace:

1. Let the coordinates of $C$ be $(x, y, z)$.
2. The formula for the centroid $G(\bar{x}, \bar{y}, \bar{z})$ of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is:$$\bar{x} = \frac{x_1 + x_2 + x_3}{3}, \quad \bar{y} = \frac{y_1 + y_2 + y_3}{3}, \quad \bar{z} = \frac{z_1 + z_2 + z_3}{3}$$
3. Substitute the known values:$$\frac{3 - 1 + x}{3} = 1 \implies 2 + x = 3 \implies x = 1$$$$\frac{-5 + 7 + y}{3} = 1 \implies 2 + y = 3 \implies y = 1$$$$\frac{7 - 6 + z}{3} = 1 \implies 1 + z = 3 \implies z = 2$$

Final Evaluation Output: The coordinates of $C$ are $(1, 1, 2)$.