3.1 Q-11

Exercise 3.1 — Question 11

Problem Statement

If $A$, $B$, and $C$ are the points $(1,-1,2)$, $(2,1,-1)$, and $(3,-1,2)$ respectively, find the vectors $\overrightarrow{AB}$, $\overrightarrow{BC}$, and $\overrightarrow{AC}$. Also verify that $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$.

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Key Formula

For any two points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$:

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

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Solution

Given: $A(1,-1,2)$, $B(2,1,-1)$, $C(3,-1,2)$

Step 1: Find $\overrightarrow{AB}$

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

Step 2: Find $\overrightarrow{BC}$

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

Step 3: Find $\overrightarrow{AC}$

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

Step 4: Verify $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$

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

$\therefore \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}✓$

This confirms the triangle law of vector addition.

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Concepts Used