3.1 Q-8 | Vectors | Mathematics 11

Exercise 3.1 — Question 8

Problem Statement

If $A = (1, 0, 2)$ , $B = (3, 1, 0)$ , $C = (5, 2, - 2)$ , show that $A$ , $B$ , and $C$ are collinear (lie on the same straight line).

Key Concepts

Vector between two points: For points $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ : $P Q = (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k}$

Condition for collinearity: Points $A$ , $B$ , $C$ are collinear if $A B$ and $A C$ are parallel, i.e., $A C = k A B$ for some scalar $k$ .

Condition for parallel vectors: Vectors $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ are parallel if: $\frac{a _{1}}{b _{1}} = \frac{a _{2}}{b _{2}} = \frac{a _{3}}{b _{3}}$

Solution

Step 1: Find $A B$ $A B = (3 - 1) \hat{i} + (1 - 0) \hat{j} + (0 - 2) \hat{k} = 2 \hat{i} + \hat{j} - 2 \hat{k}$

Step 2: Find $A C$ $A C = (5 - 1) \hat{i} + (2 - 0) \hat{j} + (- 2 - 2) \hat{k} = 4 \hat{i} + 2 \hat{j} - 4 \hat{k}$

Step 3: Check for parallelism $A C = 4 \hat{i} + 2 \hat{j} - 4 \hat{k} = 2 (2 \hat{i} + \hat{j} - 2 \hat{k}) = 2 A B$

Since $A C = 2 A B$ (i.e., $k = 2$ ), the vectors are parallel and share point $A$ .

Conclusion: $A$ , $B$ , and $C$ are collinear. $■$

General Method for Collinearity

To show three points $A$ , $B$ , $C$ are collinear:

Compute $A B$ and $A C$ .
Show $A C = k A B$ for some scalar $k \neq = 0$ .
Since both vectors share point $A$ , the three points must lie on the same line.

Step 3: Check for parallelism $A C = 4 \hat{i} + 2 \hat{j} - 4 \hat{k} = 2 (2 \hat{i} + \hat{j} - 2 \hat{k}) = 2 A B$

Since $A C = 2 A B$ (i.e., $k = 2$ ), the vectors are parallel and share point $A$ .

Conclusion: $A$ , $B$ , and $C$ are collinear. $■$

General Method for Collinearity

To show three points $A$ , $B$ , $C$ are collinear:

Compute $A B$ and $A C$ .
Show $A C = k A B$ for some scalar $k \neq = 0$ .
Since both vectors share point $A$ , the three points must lie on the same line.