Vectors in Space: 3D Coordinates, Vector Algebra, Applications

TAKEAWAYS:

• Understand 3D coordinate system (x, y, z axes) and how to locate points in space.
• Learn vector algebra (addition, subtraction, scalar multiplication) using component-wise rules.
• Master the section formula to find points dividing a line segment internally or externally.
• Check collinearity of points using vectors and ratios.
• Apply vectors to geometry problems (e.g., finding distances, angles, and areas in 3D).
• Solve real-world problems (e.g., physics, navigation, computer graphics).

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What is a Vector in 3D Space?

A vector is a quantity that has both magnitude (size) and direction. Unlike scalars (e.g., temperature, mass), vectors depend on direction. In 3D space, vectors are represented using components along the x, y, and z axes.

Position Vector

The position vector of a point $P(x,y,z)$ is the vector from the origin $O(0,0,0)$ to $P$. It is written as: $$\overrightarrow{OP}=x\hat{i}+y\hat{j}+z\hat{k}$$ where:

• $\hat{i}$ = unit vector along the x-axis,
• $\hat{j}$ = unit vector along the y-axis,
• $\hat{k}$ = unit vector along the z-axis.

Vector Between Two Points

The vector from point $A(x_1,y_1,z_1)$ to point $B(x_2,y_2,z_2)$ is: $$\overrightarrow{AB}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$$

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Vector Operations in 3D

1. Addition of Vectors

If $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, then: $$\vec{a}+\vec{b}=(a_1+b_1)\hat{i}+(a_2+b_2)\hat{j}+(a_3+b_3)\hat{k}$$

Example: Let $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=-\hat{i}+4\hat{j}+2\hat{k}$. Find $\vec{a}+\vec{b}$. Solution: $$\vec{a}+\vec{b}=(2-1)\hat{i}+(3+4)\hat{j}+(-1+2)\hat{k}=\hat{i}+7\hat{j}+\hat{k}$$

2. Subtraction of Vectors

$$\vec{a}-\vec{b}=(a_1-b_1)\hat{i}+(a_2-b_2)\hat{j}+(a_3-b_3)\hat{k}$$

Example: Using the same vectors, find $\vec{a}-\vec{b}$. Solution: $$\vec{a}-\vec{b}=(2-(-1))\hat{i}+(3-4)\hat{j}+(-1-2)\hat{k}=3\hat{i}-\hat{j}-3\hat{k}$$

3. Scalar Multiplication

Multiplying a vector $\vec{a}$ by a scalar $k$: $$k\vec{a}=(k{a}_1)\hat{i}+(k{a}_2)\hat{j}+(k{a}_3)\hat{k}$$

Example: If $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$, find $3\vec{a}$. Solution: $$3\vec{a}=6\hat{i}+9\hat{j}-3\hat{k}$$

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Section Formula in 3D

The section formula helps find a point dividing a line segment internally or externally in a given ratio.

Internal Division

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

Example: Find the coordinates of point $P$ that divides the line segment joining $A(1,2,3)$ and $B(4,5,6)$ in the ratio $2:3$. Solution: $$P=\left(\frac{2\cdot 4+3\cdot 1}{2+3},\frac{2\cdot 5+3\cdot 2}{5},\frac{2\cdot 6+3\cdot 3}{5}\right)=\left(\frac{8+3}{5},\frac{10+6}{5},\frac{12+9}{5}\right)=\left(\frac{11}{5},\frac{16}{5},\frac{21}{5}\right)$$

External Division

If point $P$ divides the line segment externally in the ratio $m:n$, then: $$P=\left(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n},\frac{m{z}_2-n{z}_1}{m-n}\right)$$

Example: Find the coordinates of point $P$ that divides the line segment joining $A(1,2,3)$ and $B(4,5,6)$ externally in the ratio $2:3$. Solution: $$P=\left(\frac{2\cdot 4-3\cdot 1}{2-3},\frac{2\cdot 5-3\cdot 2}{-1},\frac{2\cdot 6-3\cdot 3}{-1}\right)=\left(\frac{8-3}{-1},\frac{10-6}{-1},\frac{12-9}{-1}\right)=(-5,-4,-3)$$

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Collinearity of Points

Three points $A,B,$ and $C$ are collinear (lie on the same straight line) if the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are scalar multiples of each other. Mathematically, if: $$\frac{x_2-x_1}{x_3-x_1}=\frac{y_2-y_1}{y_3-y_1}=\frac{z_2-z_1}{z_3-z_1}$$ then $A,B,$ and $C$ are collinear.

Example: Check if the points $A(1,2,3)$, $B(2,4,6)$, and $C(3,6,9)$ are collinear. Solution: Calculate the ratios: $$\frac{2-1}{3-1}=\frac{1}{2},\frac{4-2}{6-2}=\frac{2}{4}=\frac{1}{2},\frac{6-3}{9-3}=\frac{3}{6}=\frac{1}{2}$$ Since all ratios are equal, the points are collinear.

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Applications of Vectors in 3D

1. Distance Between Two Points

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

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

2. Angle Between Two Vectors

The angle $\theta$ between vectors $\vec{a}$ and $\vec{b}$ is given by: $$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$$ where $\vec{a}\cdot \vec{b}$ is the dot product and $|\vec{a}|$ is the magnitude of $\vec{a}$.

Example: Find the angle between vectors $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=-\hat{i}+4\hat{j}+2\hat{k}$. Solution:

1. Dot product: $$\vec{a}\cdot \vec{b}=(2)(-1)+(3)(4)+(-1)(2)=-2+12-2=8$$
2. Magnitudes: $$|\vec{a}|=\sqrt{2^2+3^2+(-1{)}^2}=\sqrt{4+9+1}=\sqrt{14}$$ $$|\vec{b}|=\sqrt{(-1{)}^2+4^2+2^2}=\sqrt{1+16+4}=\sqrt{21}$$
3. Angle: $$\cos \theta =\frac{8}{\sqrt{14}\cdot \sqrt{21}}=\frac{8}{\sqrt{294}}⟹\theta ={\cos }^{-1}\left(\frac{8}{\sqrt{294}}\right)$$

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Comparison Table: 2D vs. 3D Vectors

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Advantages and Disadvantages of Vectors

Advantages:

1. Simplify complex problems: Vectors make it easier to solve problems involving direction and magnitude.
2. Used in physics: Helps in resolving forces, velocities, and accelerations.
3. Computer graphics: Used in 3D modeling and animations.
4. Navigation: Used in GPS and aerospace engineering.

Disadvantages:

1. Complexity in 3D: Calculations in 3D can be more involved than in 2D.
2. Visualization: Harder to visualize compared to 2D vectors.

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NEB-Style Questions

Short Answer Questions

1. Define a position vector in 3D space.
2. If $\vec{a}=3\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=-\hat{i}+4\hat{j}-2\hat{k}$, find $\vec{a}+\vec{b}$.
3. Find the coordinates of the point that divides the line segment joining $A(1,2,3)$ and $B(4,5,6)$ in the ratio $1:2$ internally.
4. Check if the points $A(1,2,3)$, $B(2,4,6)$, and $C(3,5,7)$ are collinear.

Long Answer Questions

5. Given vectors $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=-\hat{i}+4\hat{j}+2\hat{k}$:
   • Find $\vec{a}-\vec{b}$.
   • Calculate the angle between $\vec{a}$ and $\vec{b}$.
6. Find the distance between the points $A(1,-2,3)$ and $B(4,0,-1)$. Also, find the coordinates of the point that divides the line segment joining $A$ and $B$ externally in the ratio $2:3$.

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Exam Tip

1. Understand the 3D coordinate system: Always visualize vectors in 3D space.
2. Practice vector operations: Addition, subtraction, and scalar multiplication are fundamental. Solve at least 5 problems for each operation.
3. Memorize the section formula: It is frequently tested in exams. Practice both internal and external division.
4. Check collinearity carefully: Ensure you calculate the ratios correctly for all three coordinates.
5. Apply vectors to real-world problems: Questions often involve distance, angle, or division of line segments. Relate theory to practical scenarios.
6. Show all steps: In NEB exams, partial credit is given for correct steps. Always write down intermediate calculations.

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