Coordinates in Space: Points, Lines, Planes, Distances & Angles

TAKEAWAYS:

• Understand the 3D coordinate system (x, y, z axes) and how to plot points using ordered triples (x, y, z).
• Learn distance and section formula in 3D to find lengths and divide lines internally/externally.
• Master direction cosines and ratios to describe lines’ orientation in space.
• Derive equations of lines (vector/cartesian forms) and planes (general/scalar product forms).
• Calculate angles between lines/planes using dot products and direction cosines.
• Apply these concepts to real-world problems like architecture, navigation, and physics.

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1. The 3D Coordinate System

In 2D, we use two axes (x and y) to locate points. In 3D, we add a z-axis perpendicular to the xy-plane. The three axes are mutually perpendicular and intersect at the origin (0,0,0).

Key Concepts:

• Ordered Triple (x, y, z): Represents a point’s position along the x, y, and z axes.
• Octants: The 3D space is divided into 8 octants (like quadrants in 2D).
• Projection: The shadow of a point on a plane (e.g., xy-plane projection of (2,3,4) is (2,3,0)).

Worked Example 1: Plotting Points

Plot the points A(1,2,3), B(-2,4,-1), and C(0,0,5) in 3D space.

• Step 1: Draw the x, y, and z axes.
• Step 2: For A(1,2,3):
  • Move 1 unit along x, 2 units along y, and 3 units along z.
• Step 3: Repeat for B and C.

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2. Distance Between Two Points

The distance $d$ between two points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ is: $$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$$

Worked Example 2: Calculating Distance

Find the distance between A(1,2,3) and B(-2,4,-1). $$d=\sqrt{(-2-1{)}^2+(4-2{)}^2+(-1-3{)}^2}=\sqrt{9+4+16}=\sqrt{29}$$

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3. Section Formula (Internal Division)

If a point $P(x,y,z)$ divides the line joining $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $m:n$, then: $$P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n}\right)$$

Worked Example 3: Finding a Dividing Point

Find the point that divides the line joining A(1,2,3) and B(4,5,6) in the ratio 2:3. $$P=\left(\frac{2\cdot 4+3\cdot 1}{5},\frac{2\cdot 5+3\cdot 2}{5},\frac{2\cdot 6+3\cdot 3}{5}\right)=\left(\frac{11}{5},\frac{16}{5},\frac{24}{5}\right)$$

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4. Direction Cosines and Ratios

• Direction Cosines (l, m, n): Cosines of the angles a line makes with the x, y, and z axes. $$l=\cos \alpha ,m=\cos \beta ,n=\cos \gamma$$ where $\alpha ,\beta ,\gamma$ are angles with x, y, z axes.
• Property: $l^2+m^2+n^2=1$.
• Direction Ratios (a, b, c): Any three numbers proportional to direction cosines.

Worked Example 4: Finding Direction Cosines

Find the direction cosines of the line joining A(1,2,3) and B(4,5,6).

• Step 1: Find the direction ratios: $$a=4-1=3,b=5-2=3,c=6-3=3$$
• Step 2: Find the magnitude: $$\sqrt{3^2+3^2+3^2}=\sqrt{27}=3\sqrt{3}$$
• Step 3: Direction cosines: $$l=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}},m=\frac{1}{\sqrt{3}},n=\frac{1}{\sqrt{3}}$$

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5. Equation of a Line in 3D

(a) Vector Form

If a line passes through point $A(x_1,y_1,z_1)$ and has direction ratios $a,b,c$, its equation is: $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$

(b) Cartesian Form

Same as above, but written as two equations: $$\frac{x-x_1}{a}=\frac{y-y_1}{b},\frac{y-y_1}{b}=\frac{z-z_1}{c}$$

Worked Example 5: Line Equation

Find the equation of the line passing through A(1,2,3) with direction ratios 2, -1, 3. $$\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{3}$$

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6. Angle Between Two Lines

If two lines have direction ratios $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$, the angle $\theta$ between them is: $$\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\cdot \sqrt{a_2^2+b_2^2+c_2^2}}$$

Worked Example 6: Angle Between Lines

Find the angle between lines with direction ratios 1, 2, 3 and 4, -1, 2. $$\cos \theta =\frac{1\cdot 4+2\cdot (-1)+3\cdot 2}{\sqrt{1+4+9}\cdot \sqrt{16+1+4}}=\frac{4-2+6}{\sqrt{14}\cdot \sqrt{21}}=\frac{8}{\sqrt{294}}$$ $$\theta ={\cos }^{-1}\left(\frac{8}{\sqrt{294}}\right)$$

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7. Equation of a Plane

(a) General Form

$$Ax+By+Cz+D=0$$ where $(A,B,C)$ is the normal vector to the plane.

(b) Scalar Product Form

If $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ and $\vec{n}=A\hat{i}+B\hat{j}+C\hat{k}$ is the normal vector, the plane equation is: $$\vec{r}\cdot \vec{n}=d\text{(where d is a constant)}$$

(c) From Three Points

If the plane passes through points $A(x_1,y_1,z_1)$, $B(x_2,y_2,z_2)$, and $C(x_3,y_3,z_3)$, its equation is: $$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$$

Worked Example 7: Plane Equation

Find the equation of the plane passing through A(1,2,3), B(4,5,6), and C(7,8,9). $$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 3& 3& 3\\ 6& 6& 6\end{array}\right|=0$$ Since the second and third rows are proportional, the determinant is 0, meaning the points are collinear. Thus, infinitely many planes pass through them.

Alternative Approach: Use two vectors in the plane:

• $\overrightarrow{AB}=(3,3,3)$
• $\overrightarrow{AC}=(6,6,6)$ The normal vector $\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}=(0,0,0)$, confirming collinearity.

Instead, let’s take a non-collinear example: Find the plane through A(1,1,1), B(2,3,1), and C(3,1,2). $$\overrightarrow{AB}=(1,2,0),\overrightarrow{AC}=(2,0,1)$$ $$\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}=(2\cdot 1-0\cdot 0,0\cdot 2-1\cdot 1,1\cdot 0-2\cdot 2)=(2,-1,-4)$$ Plane equation: $$2(x-1)-1(y-1)-4(z-1)=0⟹2x-y-4z=-3$$

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8. Angle Between Two Planes

If two planes have normal vectors $\overrightarrow{n_1}=(A_1,B_1,C_1)$ and $\overrightarrow{n_2}=(A_2,B_2,C_2)$, the angle $\theta$ between them is: $$\cos \theta =\frac{A_1{A}_2+B_1{B}_2+C_1{C}_2}{\sqrt{A_1^2+B_1^2+C_1^2}\cdot \sqrt{A_2^2+B_2^2+C_2^2}}$$

Worked Example 8: Angle Between Planes

Find the angle between planes $2x-y+z=5$ and $x+y-2z=3$.

• Normal vectors: $\overrightarrow{n_1}=(2,-1,1)$, $\overrightarrow{n_2}=(1,1,-2)$. $$\cos \theta =\frac{2\cdot 1+(-1)\cdot 1+1\cdot (-2)}{\sqrt{4+1+1}\cdot \sqrt{1+1+4}}=\frac{2-1-2}{\sqrt{6}\cdot \sqrt{6}}=\frac{-1}{6}$$ $$\theta ={\cos }^{-1}\left(-\frac{1}{6}\right)$$

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9. Shortest Distance from a Point to a Plane

The shortest distance $D$ from point $P(x_0,y_0,z_0)$ to the plane $Ax+By+Cz+D=0$ is: $$D=\frac{|A{x}_0+B{y}_0+C{z}_0+D|}{\sqrt{A^2+B^2+C^2}}$$

Worked Example 9: Distance from Point to Plane

Find the distance from A(1,2,3) to the plane $2x-y+z=5$. $$D=\frac{|2\cdot 1-1\cdot 2+1\cdot 3-5|}{\sqrt{4+1+1}}=\frac{|2-2+3-5|}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}$$

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10. Applications of 3D Coordinates

1. Architecture: Designing 3D structures (buildings, bridges).
2. Physics: Describing motion in space (projectiles, satellites).
3. Computer Graphics: Rendering 3D objects in games and animations.
4. Navigation: GPS uses 3D coordinates to track locations.
5. Engineering: Modeling mechanical parts and robotics.

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Exam Tip

1. Memorize Formulas:
   • Distance formula, section formula, direction cosines, line/plane equations.
2. Practice Plotting Points:
   • Always visualize 3D points before calculations.
3. Check for Collinearity:
   • If three points are collinear, the plane equation is not unique.
4. Angle Calculations:
   • Use dot product for angles between lines/planes.
5. Shortest Distance:
   • Always verify if the point lies on the plane (distance = 0).
6. NEB-Style Questions:
   • Expect problems combining multiple concepts (e.g., find the angle between a line and a plane).
   • Example Question:

     "Find the equation of the plane passing through (1,2,3) and perpendicular to the line joining (4,5,6) and (7,8,9)." Solution:

     • Direction ratios of the line: $(3,3,3)$.
     • Plane equation: $3(x-1)+3(y-2)+3(z-3)=0$ → $x+y+z=6$.

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NEB Board-Style Questions

1. Short Answer:

   • Find the direction cosines of the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z-2}{4}$.

2. Long Answer:

   • Find the equation of the plane passing through (1,2,3), (4,5,6), and (7,8,9). If the points are collinear, explain why and find another plane passing through them.

3. Problem Solving:

   • A line passes through (1,2,3) and (4,5,6). Find: (i) Its direction cosines. (ii) The angle it makes with the x-axis. (iii) The equation of the plane perpendicular to this line at (1,2,3).

4. Application:

   • A cube has vertices at (0,0,0) to (1,1,1). Find the equation of the plane cutting the cube diagonally from (0,0,0) to (1,1,1).

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Summary Table