# 1.4 3D Coordinate Geometry

## Detailed Theory: Three-Dimensional Coordinate Geometry

### **1. The 3D Coordinate System**

#### **1.1 Introduction**

The three-dimensional coordinate system extends the 2D Cartesian system by adding a third axis perpendicular to both x and y axes.

#### **1.2 Components of the System**

1. **X-axis:** Horizontal axis (usually pointing right)
2. **Y-axis:** Vertical axis (usually pointing up)
3. **Z-axis:** Axis perpendicular to both x and y axes (usually pointing out of the page/screen)
4. **Origin (O):** Point (0,0,0) where all three axes intersect
5. **Coordinate Planes:**
   * **XY-plane:** z = 0
   * **YZ-plane:** x = 0
   * **ZX-plane:** y = 0
6. **Octants:** Eight regions divided by the coordinate planes
   * **Octant I:** x>0, y>0, z>0
   * **Octant II:** x<0, y>0, z>0
   * **Octant III:** x<0, y<0, z>0
   * **Octant IV:** x>0, y<0, z>0
   * **Octant V:** x>0, y>0, z<0
   * **Octant VI:** x<0, y>0, z<0
   * **Octant VII:** x<0, y<0, z<0
   * **Octant VIII:** x>0, y<0, z<0

#### **1.3 Coordinates of a Point**

A point P in 3D space is represented by an ordered triple (x, y, z):

* **x-coordinate:** Distance from YZ-plane
* **y-coordinate:** Distance from ZX-plane
* **z-coordinate:** Distance from XY-plane

**Example:** Point A(2, 3, 4) means:

* 2 units from YZ-plane (positive = right of YZ-plane)
* 3 units from ZX-plane (positive = above ZX-plane)
* 4 units from XY-plane (positive = in front of XY-plane)

***

### **2. Distance Formula in 3D**

#### **2.1 Distance Between Two Points**

The distance between points $$P(x\_1, y\_1, z\_1)$$ and $$Q(x\_2, y\_2, z\_2)$$ is: $$d = \sqrt{(x\_2 - x\_1)^2 + (y\_2 - y\_1)^2 + (z\_2 - z\_1)^2}$$

**Derivation:** Extending Pythagoras theorem to 3D.

**Example:** Distance between A(1,2,3) and B(4,6,8): $$d = \sqrt{(4-1)^2 + (6-2)^2 + (8-3)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2}$$

#### **2.2 Distance from Origin**

Distance from origin to point P(x,y,z): $$d = \sqrt{x^2 + y^2 + z^2}$$

#### **2.3 Applications**

1. **Collinearity in 3D:** Three points A, B, C are collinear if: $$AB + BC = AC$$ or $$AB + AC = BC$$ or $$AC + BC = AB$$
2. **Type of triangle in space:** Similar to 2D but with 3D distances

***

### **3. Section Formula in 3D**

#### **3.1 Internal Division**

If point P divides line segment AB internally in ratio m:n, where A(x₁,y₁,z₁) and B(x₂,y₂,z₂), then: $$P\left(\frac{mx\_2 + nx\_1}{m+n}, \frac{my\_2 + ny\_1}{m+n}, \frac{mz\_2 + nz\_1}{m+n}\right)$$

**Example:** Point dividing A(1,2,3) and B(4,6,8) in ratio 2:3 internally: $$P\left(\frac{2\times4 + 3\times1}{2+3}, \frac{2\times6 + 3\times2}{2+3}, \frac{2\times8 + 3\times3}{2+3}\right) = \left(\frac{8+3}{5}, \frac{12+6}{5}, \frac{16+9}{5}\right) = \left(\frac{11}{5}, \frac{18}{5}, 5\right)$$

#### **3.2 External Division**

If point P divides AB externally in ratio m:n (m≠n), then: $$P\left(\frac{mx\_2 - nx\_1}{m-n}, \frac{my\_2 - ny\_1}{m-n}, \frac{mz\_2 - nz\_1}{m-n}\right)$$

#### **3.3 Midpoint Formula**

Midpoint of A(x₁,y₁,z₁) and B(x₂,y₂,z₂): $$M\left(\frac{x\_1+x\_2}{2}, \frac{y\_1+y\_2}{2}, \frac{z\_1+z\_2}{2}\right)$$

#### **3.4 Centroid of Tetrahedron**

For tetrahedron with vertices A(x₁,y₁,z₁), B(x₂,y₂,z₂), C(x₃,y₃,z₃), D(x₄,y₄,z₄): $$G\left(\frac{x\_1+x\_2+x\_3+x\_4}{4}, \frac{y\_1+y\_2+y\_3+y\_4}{4}, \frac{z\_1+z\_2+z\_3+z\_4}{4}\right)$$

#### **3.5 Centroid of Triangle in 3D**

For triangle with vertices A(x₁,y₁,z₁), B(x₂,y₂,z₂), C(x₃,y₃,z₃): $$G\left(\frac{x\_1+x\_2+x\_3}{3}, \frac{y\_1+y\_2+y\_3}{3}, \frac{z\_1+z\_2+z\_3}{3}\right)$$

***

### **4. Direction Cosines and Direction Ratios**

#### **4.1 Direction Cosines**

For a line making angles α, β, γ with positive x, y, z axes respectively:

* $$\cos\alpha = l$$
* $$\cos\beta = m$$
* $$\cos\gamma = n$$

**Properties:**

1. $$l^2 + m^2 + n^2 = 1$$
2. Direction cosines of x-axis: (1,0,0)
3. Direction cosines of y-axis: (0,1,0)
4. Direction cosines of z-axis: (0,0,1)

#### **4.2 Direction Ratios**

Three numbers a, b, c proportional to direction cosines: $$\frac{l}{a} = \frac{m}{b} = \frac{n}{c} = \frac{1}{\sqrt{a^2 + b^2 + c^2}}$$

**Relation between direction cosines and ratios:** $$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$$

#### **4.3 Direction Cosines of Line Through Two Points**

For line through P(x₁,y₁,z₁) and Q(x₂,y₂,z₂), direction ratios are: $$(x\_2-x\_1, y\_2-y\_1, z\_2-z\_1)$$

Direction cosines are: $$\left(\frac{x\_2-x\_1}{d}, \frac{y\_2-y\_1}{d}, \frac{z\_2-z\_1}{d}\right)$$ where $$d = \sqrt{(x\_2-x\_1)^2 + (y\_2-y\_1)^2 + (z\_2-z\_1)^2}$$

***

### **5. Angle Between Two Lines**

#### **5.1 Angle Using Direction Cosines**

If two lines have direction cosines (l₁,m₁,n₁) and (l₂,m₂,n₂), the angle θ between them is: $$\cos\theta = l\_1l\_2 + m\_1m\_2 + n\_1n\_2$$

**Special Cases:**

1. **Parallel lines:** $$l\_1 = l\_2, m\_1 = m\_2, n\_1 = n\_2$$
2. **Perpendicular lines:** $$l\_1l\_2 + m\_1m\_2 + n\_1n\_2 = 0$$

#### **5.2 Angle Using Direction Ratios**

If two lines have direction ratios (a₁,b₁,c₁) and (a₂,b₂,c₂), then: $$\cos\theta = \frac{a\_1a\_2 + b\_1b\_2 + c\_1c\_2}{\sqrt{a\_1^2 + b\_1^2 + c\_1^2}\sqrt{a\_2^2 + b\_2^2 + c\_2^2}}$$

**Example:** Find angle between lines with direction ratios (1,2,3) and (4,5,6): $$\cos\theta = \frac{1\times4 + 2\times5 + 3\times6}{\sqrt{1^2+2^2+3^2}\sqrt{4^2+5^2+6^2}} = \frac{4+10+18}{\sqrt{14}\sqrt{77}} = \frac{32}{\sqrt{1078}} \approx 0.9746$$ $$\theta \approx \cos^{-1}(0.9746) \approx 12.9^\circ$$

***

### **6. Projection of a Line Segment**

#### **6.1 Projection on Coordinate Axes**

Projection of line segment PQ on:

* **x-axis:** $$|x\_2 - x\_1|$$
* **y-axis:** $$|y\_2 - y\_1|$$
* **z-axis:** $$|z\_2 - z\_1|$$

#### **6.2 Projection on a Line with Given Direction Cosines**

Projection of PQ with direction cosines (l,m,n): $$\text{Projection} = |(x\_2-x\_1)l + (y\_2-y\_1)m + (z\_2-z\_1)n|$$

**Example:** Projection of segment from (1,2,3) to (4,6,8) on line with direction cosines (1/√3, 1/√3, 1/√3): $$\text{Proj} = \left|(4-1)\frac{1}{\sqrt{3}} + (6-2)\frac{1}{\sqrt{3}} + (8-3)\frac{1}{\sqrt{3}}\right| = \left|\frac{3+4+5}{\sqrt{3}}\right| = \frac{12}{\sqrt{3}} = 4\sqrt{3}$$

***

### **7. Equations of Lines in 3D**

#### **7.1 Vector Form**

Line through point with position vector $$\vec{a}$$ and parallel to vector $$\vec{b}$$: $$\vec{r} = \vec{a} + \lambda\vec{b}$$ where λ is a scalar parameter.

#### **7.2 Cartesian Form**

**a) Symmetric Form (Two-Point Form)**

Line through (x₁,y₁,z₁) and (x₂,y₂,z₂): $$\frac{x-x\_1}{x\_2-x\_1} = \frac{y-y\_1}{y\_2-y\_1} = \frac{z-z\_1}{z\_2-z\_1}$$

**b) Symmetric Form (Direction Ratios)**

Line through (x₁,y₁,z₁) with direction ratios a,b,c: $$\frac{x-x\_1}{a} = \frac{y-y\_1}{b} = \frac{z-z\_1}{c}$$

**c) Parametric Form**

Line through (x₁,y₁,z₁) with direction ratios a,b,c:

$$
\begin{cases}
x = x\_1 + aλ \\
y = y\_1 + bλ \\
z = z\_1 + cλ
\end{cases}
$$

where λ is parameter.

#### **7.3 Conversion Between Forms**

**Example:** Convert line $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ to parametric form: $$x = 1 + 2λ, \quad y = 2 + 3λ, \quad z = 3 + 4λ$$

***

### **8. Angle Between Lines**

#### **8.1 Formula Using Direction Ratios**

For lines: $$L\_1: \frac{x-x\_1}{a\_1} = \frac{y-y\_1}{b\_1} = \frac{z-z\_1}{c\_1}$$ $$L\_2: \frac{x-x\_2}{a\_2} = \frac{y-y\_2}{b\_2} = \frac{z-z\_2}{c\_2}$$

Angle θ between them: $$\cos\theta = \frac{a\_1a\_2 + b\_1b\_2 + c\_1c\_2}{\sqrt{a\_1^2+b\_1^2+c\_1^2}\sqrt{a\_2^2+b\_2^2+c\_2^2}}$$

#### **8.2 Special Cases**

1. **Parallel lines:** $$\frac{a\_1}{a\_2} = \frac{b\_1}{b\_2} = \frac{c\_1}{c\_2}$$
2. **Perpendicular lines:** $$a\_1a\_2 + b\_1b\_2 + c\_1c\_2 = 0$$

***

### **9. Shortest Distance Between Lines**

#### **9.2 Distance Between Parallel Lines**

Given two parallel lines:

Line 1: $$\frac{x - x\_1}{a} = \frac{y - y\_1}{b} = \frac{z - z\_1}{c}$$

Line 2: $$\frac{x - x\_2}{a} = \frac{y - y\_2}{b} = \frac{z - z\_2}{c}$$

**Vector Form Formula:** The distance between these parallel lines is: $$d = \frac{|(\vec{a}\_2 - \vec{a}\_1) \times \vec{b}|}{|\vec{b}|}$$

Where:

* $$\vec{a}\_1 = (x\_1, y\_1, z\_1)$$ (position vector of point on line 1)
* $$\vec{a}\_2 = (x\_2, y\_2, z\_2)$$ (position vector of point on line 2)
* $$\vec{b} = (a, b, c)$$ (direction vector of both lines)

**Cartesian Formula:** The distance can also be calculated using: $$d = \sqrt{\frac{D\_1^2 + D\_2^2 + D\_3^2}{a^2 + b^2 + c^2}}$$

Where: $$D\_1 = \begin{vmatrix} y\_2 - y\_1 & z\_2 - z\_1 \ b & c \end{vmatrix}$$ $$D\_2 = \begin{vmatrix} z\_2 - z\_1 & x\_2 - x\_1 \ c & a \end{vmatrix}$$ $$D\_3 = \begin{vmatrix} x\_2 - x\_1 & y\_2 - y\_1 \ a & b \end{vmatrix}$$

**Expanded Form:** $$d = \frac{\sqrt{(b(z\_2 - z\_1) - c(y\_2 - y\_1))^2 + (c(x\_2 - x\_1) - a(z\_2 - z\_1))^2 + (a(y\_2 - y\_1) - b(x\_2 - x\_1))^2}}{\sqrt{a^2 + b^2 + c^2}}$$

**Example:** Find the distance between the parallel lines:

Line 1: $$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$$

Line 2: $$\frac{x - 4}{2} = \frac{y - 5}{3} = \frac{z - 6}{4}$$

**Solution:**

Given:

* $$x\_1 = 1, y\_1 = 2, z\_1 = 3$$
* $$x\_2 = 4, y\_2 = 5, z\_2 = 6$$
* $$a = 2, b = 3, c = 4$$

**Step 1: Calculate D₁, D₂, D₃**

$$D\_1 = \begin{vmatrix} y\_2 - y\_1 & z\_2 - z\_1 \ b & c \end{vmatrix} = \begin{vmatrix} 3 & 3 \ 3 & 4 \end{vmatrix} = (3 \times 4) - (3 \times 3) = 12 - 9 = 3$$

$$D\_2 = \begin{vmatrix} z\_2 - z\_1 & x\_2 - x\_1 \ c & a \end{vmatrix} = \begin{vmatrix} 3 & 3 \ 4 & 2 \end{vmatrix} = (3 \times 2) - (3 \times 4) = 6 - 12 = -6$$

$$D\_3 = \begin{vmatrix} x\_2 - x\_1 & y\_2 - y\_1 \ a & b \end{vmatrix} = \begin{vmatrix} 3 & 3 \ 2 & 3 \end{vmatrix} = (3 \times 3) - (3 \times 2) = 9 - 6 = 3$$

**Step 2: Calculate the distance**

$$d = \sqrt{\frac{D\_1^2 + D\_2^2 + D\_3^2}{a^2 + b^2 + c^2}} = \sqrt{\frac{3^2 + (-6)^2 + 3^2}{2^2 + 3^2 + 4^2}}$$

$$d = \sqrt{\frac{9 + 36 + 9}{4 + 9 + 16}} = \sqrt{\frac{54}{29}} = \frac{3\sqrt{6}}{\sqrt{29}}$$

**Final Answer:** The distance between the parallel lines is $$\frac{3\sqrt{6}}{\sqrt{29}}$$ units.

#### **9.3 Distance Between Skew Lines**

For lines: $$\vec{r} = \vec{a}\_1 + \lambda\vec{b}\_1$$ and $$\vec{r} = \vec{a}\_2 + \mu\vec{b}\_2$$

Shortest distance $$d = \frac{|(\vec{a}\_2 - \vec{a}\_1) \cdot (\vec{b}\_1 \times \vec{b}\_2)|}{|\vec{b}\_1 \times \vec{b}\_2|}$$

**Cartesian formula for lines:** $$L\_1: \frac{x - x\_1}{a\_1} = \frac{y - y\_1}{b\_1} = \frac{z - z\_1}{c\_1}$$ $$L\_2: \frac{x - x\_2}{a\_2} = \frac{y - y\_2}{b\_2} = \frac{z - z\_2}{c\_2}$$

Distance $$d = \frac{ \begin{vmatrix} x\_2 - x\_1 & y\_2 - y\_1 & z\_2 - z\_1 \ a\_1 & b\_1 & c\_1 \ a\_2 & b\_2 & c\_2 \end{vmatrix} }{ \sqrt{ \begin{vmatrix} b\_1 & c\_1 \ b\_2 & c\_2 \end{vmatrix}^2 + \begin{vmatrix} c\_1 & a\_1 \ c\_2 & a\_2 \end{vmatrix}^2 + \begin{vmatrix} a\_1 & b\_1 \ a\_2 & b\_2 \end{vmatrix}^2 }}$$

***

### **10. Equations of Planes**

#### **10.1 General Equation**

The general equation of a plane is: $$Ax + By + Cz + D = 0$$ where A, B, C are not all zero.

#### **10.2 Normal Form**

Plane with perpendicular distance p from origin and direction cosines of normal (l,m,n): $$lx + my + nz = p$$

#### **10.3 Intercept Form**

Plane making intercepts a, b, c on x, y, z axes: $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

#### **10.4 Three-Point Form**

Plane through points (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃):

$$
\begin{vmatrix}
x - x\_1 & y - y\_1 & z - z\_1 \\
x\_2 - x\_1 & y\_2 - y\_1 & z\_2 - z\_1 \\
x\_3 - x\_1 & y\_3 - y\_1 & z\_3 - z\_1
\end{vmatrix} = 0
$$

**Alternative form:**

$$
\begin{vmatrix}
x & y & z & 1 \\
x\_1 & y\_1 & z\_1 & 1 \\
x\_2 & y\_2 & z\_2 & 1 \\
x\_3 & y\_3 & z\_3 & 1
\end{vmatrix} = 0
$$

#### **10.5 Equation from Point and Normal**

Plane through (x₁,y₁,z₁) with normal having direction ratios (A,B,C): $$A(x - x\_1) + B(y - y\_1) + C(z - z\_1) = 0$$

#### **10.6 Equation from Intercepts**

If plane cuts x-axis at (a,0,0), y-axis at (0,b,0), z-axis at (0,0,c): $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

***

### **11. Angle Between Planes**

#### **11.1 Using Normal Vectors**

For planes: $$A\_1x + B\_1y + C\_1z + D\_1 = 0$$ and $$A\_2x + B\_2y + C\_2z + D\_2 = 0$$

Angle θ between them is angle between their normals: $$\cos\theta = \frac{|A\_1A\_2 + B\_1B\_2 + C\_1C\_2|}{\sqrt{A\_1^2 + B\_1^2 + C\_1^2}\sqrt{A\_2^2 + B\_2^2 + C\_2^2}}$$

#### **11.2 Special Cases**

1. **Parallel planes:** $$\frac{A\_1}{A\_2} = \frac{B\_1}{B\_2} = \frac{C\_1}{C\_2} \neq \frac{D\_1}{D\_2}$$
2. **Perpendicular planes:** $$A\_1A\_2 + B\_1B\_2 + C\_1C\_2 = 0$$

***

### **12. Distance of a Point from a Plane**

#### **12.1 Distance Formula**

Distance from point P(x₁,y₁,z₁) to plane $$Ax + By + Cz + D = 0$$: $$d = \frac{|Ax\_1 + By\_1 + Cz\_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

**Example:** Distance from (1,2,3) to plane $$2x - 3y + 6z - 5 = 0$$: $$d = \frac{|2(1) - 3(2) + 6(3) - 5|}{\sqrt{2^2 + (-3)^2 + 6^2}} = \frac{|2 - 6 + 18 - 5|}{\sqrt{4 + 9 + 36}} = \frac{9}{\sqrt{49}} = \frac{9}{7}$$

#### **12.2 Distance Between Parallel Planes**

Distance between planes $$Ax + By + Cz + D\_1 = 0$$ and $$Ax + By + Cz + D\_2 = 0$$: $$d = \frac{|D\_1 - D\_2|}{\sqrt{A^2 + B^2 + C^2}}$$

***

### **13. Angle Between Line and Plane**

#### **13.1 Formula**

For line with direction ratios (a,b,c) and plane with normal ratios (A,B,C), if θ is angle between line and plane, and φ is angle between line and normal: $$\sin\theta = \cos\phi = \frac{|Aa + Bb + Cc|}{\sqrt{A^2 + B^2 + C^2}\sqrt{a^2 + b^2 + c^2}}$$

**Alternative:** If line makes angle α with plane, and direction cosines of line are (l,m,n) while normal to plane has direction cosines (L,M,N): $$\sin\alpha = |lL + mM + nN|$$

#### **13.2 Special Cases**

1. **Line parallel to plane:** $$Aa + Bb + Cc = 0$$
2. **Line perpendicular to plane:** $$\frac{a}{A} = \frac{b}{B} = \frac{c}{C}$$

***

### **14. Sphere**

#### **14.1 Standard Equations**

**a) Center at Origin, Radius r**

$$x^2 + y^2 + z^2 = r^2$$

**b) Center at (h,k,l), Radius r**

$$(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$$

**c) General Form**

$$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$$ where center = $$(-u, -v, -w)$$ and radius = $$\sqrt{u^2 + v^2 + w^2 - d}$$

**Conditions:**

* Real sphere: $$u^2 + v^2 + w^2 - d > 0$$
* Point sphere: $$u^2 + v^2 + w^2 - d = 0$$
* Imaginary sphere: $$u^2 + v^2 + w^2 - d < 0$$

#### **14.2 Equation with Diameter Ends**

If (x₁,y₁,z₁) and (x₂,y₂,z₂) are endpoints of diameter: $$(x - x\_1)(x - x\_2) + (y - y\_1)(y - y\_2) + (z - z\_1)(z - z\_2) = 0$$

#### **14.3 Plane Section of a Sphere**

Intersection of sphere and plane is a circle.

***

### **15. Cylinder**

#### **15.1 Right Circular Cylinder**

Cylinder with axis as z-axis and radius r: $$x^2 + y^2 = r^2$$

#### **15.2 General Cylinder**

Cylinder with generators parallel to line with direction ratios (l,m,n) and guiding curve f(x,y)=0 in plane z=0: Equation obtained by eliminating λ from: $$x = x\_1 + l\lambda, \quad y = y\_1 + m\lambda, \quad z = n\lambda$$ where (x₁,y₁) satisfies f(x,y)=0.

***

### **16. Cone**

#### **16.1 Right Circular Cone**

Cone with vertex at origin, axis along z-axis, and semi-vertical angle α: $$x^2 + y^2 = z^2\tan^2\alpha$$

#### **16.2 General Cone**

Cone with vertex at (x₁,y₁,z₁) and guiding curve f(x,y)=0 in plane z=0: Equation obtained by eliminating λ from: $$x = x\_1 + l\lambda, \quad y = y\_1 + m\lambda, \quad z = z\_1 + n\lambda$$ where (l,m,n) are direction ratios of generator.

***

### **17. Solved Examples**

#### **Example 1:** Distance and Section Formula

Find coordinates of point dividing line joining A(1,2,3) and B(4,6,8) in ratio 2:1 internally and externally.

**Solution:**

**Internal division (m:n = 2:1):** $$P\left(\frac{2\times4 + 1\times1}{2+1}, \frac{2\times6 + 1\times2}{2+1}, \frac{2\times8 + 1\times3}{2+1}\right) = \left(\frac{8+1}{3}, \frac{12+2}{3}, \frac{16+3}{3}\right) = \left(3, \frac{14}{3}, \frac{19}{3}\right)$$

**External division (m:n = 2:1):** $$P\left(\frac{2\times4 - 1\times1}{2-1}, \frac{2\times6 - 1\times2}{2-1}, \frac{2\times8 - 1\times3}{2-1}\right) = \left(\frac{8-1}{1}, \frac{12-2}{1}, \frac{16-3}{1}\right) = (7, 10, 13)$$

#### **Example 2:** Equation of Plane

Find equation of plane through points A(1,1,1), B(1,-1,1), C(-1,-1,1).

**Solution:** Using three-point form:

$$
\begin{vmatrix}
x - 1 & y - 1 & z - 1 \\
1 - 1 & -1 - 1 & 1 - 1 \\
-1 - 1 & -1 - 1 & 1 - 1
\end{vmatrix} = 0
$$

Simplify:

$$
\begin{vmatrix}
x - 1 & y - 1 & z - 1 \\
0 & -2 & 0 \\
-2 & -2 & 0
\end{vmatrix} = 0
$$

Expand determinant: $$(x-1)\[(-2)(0) - (0)(-2)] - (y-1)\[(0)(0) - (0)(-2)] + (z-1)\[(0)(-2) - (-2)(-2)] = 0$$ $$0 - 0 + (z-1)\[0 - 4] = 0$$ $$-4(z-1) = 0$$ $$z = 1$$

So the plane is $$z = 1$$ (horizontal plane).

#### **Example 3:** Angle Between Line and Plane

Find angle between line $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ and plane $$x + 2y + 3z = 4$$.

**Solution:** Direction ratios of line: (2,3,4) Normal to plane: (1,2,3)

$$\sin\theta = \frac{|2\times1 + 3\times2 + 4\times3|}{\sqrt{2^2+3^2+4^2}\sqrt{1^2+2^2+3^2}} = \frac{|2+6+12|}{\sqrt{4+9+16}\sqrt{1+4+9}} = \frac{20}{\sqrt{29}\sqrt{14}} = \frac{20}{\sqrt{406}}$$

$$\theta = \sin^{-1}\left(\frac{20}{\sqrt{406}}\right) \approx \sin^{-1}(0.992) \approx 82.8^\circ$$

#### **Example 4:** Distance Between Skew Lines

Find shortest distance between lines: $$L\_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ $$L\_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$

**Solution:** Given:

* For L₁: $$x\_1=1, y\_1=2, z\_1=3, a\_1=2, b\_1=3, c\_1=4$$
* For L₂: $$x\_2=2, y\_2=4, z\_2=5, a\_2=3, b\_2=4, c\_2=5$$

**Step 1: Calculate the determinant in numerator:**

$$
\begin{vmatrix}
x\_2-x\_1 & y\_2-y\_1 & z\_2-z\_1 \\
a\_1 & b\_1 & c\_1 \\
a\_2 & b\_2 & c\_2
\end{vmatrix} =
\begin{vmatrix}
1 & 2 & 2 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{vmatrix}
$$

Expanding: $$= 1\begin{vmatrix}3 & 4 \ 4 & 5\end{vmatrix} - 2\begin{vmatrix}2 & 4 \ 3 & 5\end{vmatrix} + 2\begin{vmatrix}2 & 3 \ 3 & 4\end{vmatrix}$$ $$= 1(15-16) - 2(10-12) + 2(8-9)$$ $$= 1(-1) - 2(-2) + 2(-1) = -1 + 4 - 2 = 1$$

**Step 2: Calculate the denominator:**

First, calculate the three 2×2 determinants:

$$\begin{vmatrix} b\_1 & c\_1 \ b\_2 & c\_2 \end{vmatrix} = \begin{vmatrix} 3 & 4 \ 4 & 5 \end{vmatrix} = 15 - 16 = -1$$

$$\begin{vmatrix} c\_1 & a\_1 \ c\_2 & a\_2 \end{vmatrix} = \begin{vmatrix} 4 & 2 \ 5 & 3 \end{vmatrix} = 12 - 10 = 2$$

$$\begin{vmatrix} a\_1 & b\_1 \ a\_2 & b\_2 \end{vmatrix} = \begin{vmatrix} 2 & 3 \ 3 & 4 \end{vmatrix} = 8 - 9 = -1$$

Denominator = $$\sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$$

**Step 3: Calculate distance:**

$$d = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$$

**Final Answer:** The shortest distance between the skew lines is $$\frac{1}{\sqrt{6}}$$ units.

***

### **18. Practice Tips for Exams**

1. **3D Distance:** $$d = \sqrt{(x\_2-x\_1)^2 + (y\_2-y\_1)^2 + (z\_2-z\_1)^2}$$
2. **Section Formula:** Internal: $$\left(\frac{mx\_2+nx\_1}{m+n}, \frac{my\_2+ny\_1}{m+n}, \frac{mz\_2+nz\_1}{m+n}\right)$$
3. **Direction Cosines:** $$l^2+m^2+n^2=1$$
4. **Angle between lines:** $$\cos\theta = l\_1l\_2+m\_1m\_2+n\_1n\_2$$
5. **Plane equation:** General form: $$Ax+By+Cz+D=0$$
6. **Distance point to plane:** $$d = \frac{|Ax\_1+By\_1+Cz\_1+D|}{\sqrt{A^2+B^2+C^2}}$$
7. **Angle line-plane:** $$\sin\theta = \frac{|Aa+Bb+Cc|}{\sqrt{A^2+B^2+C^2}\sqrt{a^2+b^2+c^2}}$$
8. **Sphere:** Center $$(-u,-v,-w)$$, radius $$\sqrt{u^2+v^2+w^2-d}$$
9. **Visualization:** Always try to visualize in 3D
10. **Check octants:** Consider signs of coordinates
11. **Cross product:** Useful for normals and distances
12. **Determinants:** Use for area/volume calculations

This comprehensive theory covers all aspects of 3D Coordinate Geometry with detailed explanations and examples, providing complete preparation for the entrance examination.