3.1 Q-12

Exercise 3.1 — Question 12

Problem Statement

If $A=(1,0,2)$, $B=(2,1,3)$, $C=(4,3,5)$, show that $A$, $B$, and $C$ are collinear.

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Key Concept: Collinearity of Three Points

Three points $A$, $B$, $C$ are collinear (lie on the same straight line) if and only if the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel — i.e., one is a scalar multiple of the other.

$\overrightarrow{AB}\parallel \overrightarrow{AC}⟺\overrightarrow{AC}=k\overrightarrow{AB}\text{ for some scalar }k$

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Finding a Vector Between Two Points

Given points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$:

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

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Worked Solution

Given $A=(1,0,2)$, $B=(2,1,3)$, $C=(4,3,5)$:

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

Step 2: Find $\overrightarrow{AC}$: $\overrightarrow{AC}=(4-1)\hat{i}+(3-0)\hat{j}+(5-2)\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}$

Step 3: Check for parallelism: $\overrightarrow{AC}=3(\hat{i}+\hat{j}+\hat{k})=3\overrightarrow{AB}$

Since $\overrightarrow{AC}=3\overrightarrow{AB}$, the vectors are parallel and share point $A$.

$\therefore A,B,C\text{ are collinear.}■$

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Condition for Parallel Vectors

Two vectors $\mathbf{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\mathbf{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ are parallel if:

$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}$

or equivalently, $\mathbf{a}=k\mathbf{b}$ for some scalar $k\in \mathbb{R}$.

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