3.1 Q-13

Exercise 3.1 — Question 13

Problem

Find the vector $\overrightarrow{PQ}$ if $P=(2,-1,3)$ and $Q=(5,4,-2)$. Also determine whether $\overrightarrow{PQ}$ is parallel to $\vec{r}=3\hat{i}+5\hat{j}-5\hat{k}$.

---

Key Concepts

Vector between two points: Given points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, the vector from $P$ to $Q$ is:

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

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 $k\ne 0$.

---

Solution

Step 1: Find $\overrightarrow{PQ}$

Given $P(2,-1,3)$ and $Q(5,4,-2)$:

$\overrightarrow{PQ}=(5-2)\hat{i}+(4-(-1))\hat{j}+(-2-3)\hat{k}$

$\overrightarrow{PQ}=3\hat{i}+5\hat{j}-5\hat{k}$

Step 2: Check for parallelism with $\vec{r}=3\hat{i}+5\hat{j}-5\hat{k}$

Compare the ratios of corresponding components:

$\frac{3}{3}=\frac{5}{5}=\frac{-5}{-5}=1$

Since all ratios are equal (and $k=1$), we have $\overrightarrow{PQ}=1\cdot \vec{r}$.

$\boxed{\overrightarrow{PQ}=3\hat{i}+5\hat{j}-5\hat{k}\parallel \vec{r}}$

In fact, $\overrightarrow{PQ}=\vec{r}$, so they are not just parallel but identical vectors.

---

Summary