3.1 Q-6

Exercise 3.1 — Question 6

Problem Statement

Find the vector $\overrightarrow{PQ}$ and its magnitude, given the coordinates of points $P$ and $Q$ in space. Also determine whether the given vectors are parallel.

---

Key Concepts

Vector Between Two Points in Space

Given two points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ in 3D space, the vector from $P$ to $Q$ is:

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

This is obtained by subtracting the position vector of the initial point from the position vector of the terminal point.

Magnitude of a Vector

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

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

This extends the Pythagorean theorem to three dimensions.

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:

$\mathbf{a}=k\mathbf{b}\text{for some scalar }k$

Equivalently, their components are proportional:

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

---

Worked Example

Given: $P(1,-2,3)$ and $Q(4,0,-1)$

Step 1: Find $\overrightarrow{PQ}$

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

Step 2: Find the magnitude

$|\overrightarrow{PQ}|=\sqrt{(3{)}^2+(2{)}^2+(-4{)}^2}=\sqrt{9+4+16}=\sqrt{29}$

---

Checking for Parallel Vectors

Example: Are $\mathbf{a}=2\hat{i}-4\hat{j}+6\hat{k}$ and $\mathbf{b}=\hat{i}-2\hat{j}+3\hat{k}$ parallel?

Check if $\mathbf{a}=k\mathbf{b}$:

$\frac{2}{1}=\frac{-4}{-2}=\frac{6}{3}=2$

Since all ratios equal $2$, we have $\mathbf{a}=2\mathbf{b}$. Therefore, $\mathbf{a}$ and $\mathbf{b}$ are parallel.