3.1 Q-15

Exercise 3.1 — Question 15

Problem Statement

If $\vec{a}$, $\vec{b}$, $\vec{c}$ are vectors, prove the following properties of vector addition:

1. Commutative Law: $\vec{a}+\vec{b}=\vec{b}+\vec{a}$
2. Associative Law: $(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$
3. Identity (Null Vector): $\vec{a}+\vec{0}=\vec{a}$
4. Additive Inverse: $\vec{a}+(-\vec{a})=\vec{0}$

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Key Concepts Required

1. Rectangular Coordinate System in Space (M-11-B-01)

In 3D space, every point is located by an ordered triple $(x,y,z)$ measured along three mutually perpendicular axes: the $x$-axis, $y$-axis, and $z$-axis. The position vector of point $P(x_1,y_1,z_1)$ from the origin $O$ is:

$\overrightarrow{OP}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$

2. Unit Vectors and Components (M-11-B-02)

The unit vectors $\hat{i}$, $\hat{j}$, $\hat{k}$ point along the positive $x$-, $y$-, and $z$-axes respectively, each with magnitude 1. Any vector $\vec{v}$ in space can be written as:

$\vec{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$

where $v_1$, $v_2$, $v_3$ are the scalar components.

3. Vector Between Two Points (M-11-B-04)

If $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ are two points in space, then:

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

4. Magnitude of a Vector (M-11-B-03)

For $\vec{v}=a\hat{i}+b\hat{j}+c\hat{k}$:

$|\vec{v}|=\sqrt{a^2+b^2+c^2}$

5. Condition for Parallel Vectors (M-11-B-04)

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

$\vec{a}=k\vec{b}\text{for some scalar }k\ne 0$

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

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Proofs of Vector Addition Properties (M-11-B-05)

Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$.

Proof 1: Commutative Law — $\vec{a}+\vec{b}=\vec{b}+\vec{a}$

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

Since addition of real numbers is commutative ($a_i+b_i=b_i+a_i$):

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

Proof 2: Associative Law — $(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$

$(\vec{a}+\vec{b})+\vec{c}=[(a_1+b_1)+c_1]\hat{i}+[(a_2+b_2)+c_2]\hat{j}+[(a_3+b_3)+c_3]\hat{k}$

By associativity of real number addition:

$=[a_1+(b_1+c_1)]\hat{i}+[a_2+(b_2+c_2)]\hat{j}+[a_3+(b_3+c_3)]\hat{k}=\vec{a}+(\vec{b}+\vec{c})✓$

Proof 3: Null Vector as Identity — $\vec{a}+\vec{0}=\vec{a}$

The null vector is $\vec{0}=0\hat{i}+0\hat{j}+0\hat{k}$.

$\vec{a}+\vec{0}=(a_1+0)\hat{i}+(a_2+0)\hat{j}+(a_3+0)\hat{k}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}=\vec{a}✓$

Proof 4: Additive Inverse — $\vec{a}+(-\vec{a})=\vec{0}$

The additive inverse of $\vec{a}$ is $-\vec{a}=-a_1\hat{i}-a_2\hat{j}-a_3\hat{k}$.

$\vec{a}+(-\vec{a})=(a_1-a_1)\hat{i}+(a_2-a_2)\hat{j}+(a_3-a_3)\hat{k}=0\hat{i}+0\hat{j}+0\hat{k}=\vec{0}✓$

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Worked Example: Finding $\overrightarrow{PQ}$ and $|\overrightarrow{PQ}|$

Given: $P(1,2,3)$ and $Q(4,6,8)$

Step 1: Apply the formula: $\overrightarrow{PQ}=(4-1)\hat{i}+(6-2)\hat{j}+(8-3)\hat{k}=3\hat{i}+4\hat{j}+5\hat{k}$

Step 2: Find the magnitude: $|\overrightarrow{PQ}|=\sqrt{3^2+4^2+5^2}=\sqrt{9+16+25}=\sqrt{50}=5\sqrt{2}$

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Summary Table

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