3.1 Q-7

Exercise 3.1 — Question 7

Problem Statement

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

$\overrightarrow{AC}=k\overrightarrow{AB}\text{for some scalar }k\ne 0$

In component form, two vectors $\mathbf{u}=u_1\hat{i}+u_2\hat{j}+u_3\hat{k}$ and $\mathbf{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$ are parallel when:

$\frac{u_1}{v_1}=\frac{u_2}{v_2}=\frac{u_3}{v_3}$

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

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 $\overrightarrow{AC}$:

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

Step 3: Check if $\overrightarrow{AC}=k\overrightarrow{AB}$:

$2\hat{i}+2\hat{j}+2\hat{k}=2(\hat{i}+\hat{j}+\hat{k})$

So $k=2$. Since $\overrightarrow{AC}=2\overrightarrow{AB}$, the vectors are parallel and share point $A$.

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

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Summary of Method