3.1 Q-14

Exercise 3.1 — Question 14

Problem Statement

Find the vector $\overrightarrow{PQ}$ and its magnitude, given the points $P(1,2,-3)$ and $Q(4,-1,5)$ in space. Also determine whether $\overrightarrow{PQ}$ and $\overrightarrow{QP}$ are parallel.

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Key Concepts

Vector between two points in 3D space:

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

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

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

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

Parallel vectors: Two vectors $\vec{a}$ and $\vec{b}$ are parallel if $\vec{a}=k\vec{b}$ for some scalar $k$, i.e., their components are proportional:

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

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Solution

Step 1: Find $\overrightarrow{PQ}$

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

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

$\overrightarrow{PQ}=3\hat{i}-3\hat{j}+8\hat{k}$

Step 2: Find $|\overrightarrow{PQ}|$

$|\overrightarrow{PQ}|=\sqrt{3^2+(-3{)}^2+8^2}=\sqrt{9+9+64}=\sqrt{82}$

Step 3: Find $\overrightarrow{QP}$

$\overrightarrow{QP}=-\overrightarrow{PQ}=-3\hat{i}+3\hat{j}-8\hat{k}$

Step 4: Check if $\overrightarrow{PQ}$ and $\overrightarrow{QP}$ are parallel

Since $\overrightarrow{QP}=-1\cdot \overrightarrow{PQ}$, they are parallel (antiparallel — same line, opposite directions). The scalar $k=-1$.

$\frac{-3}{3}=\frac{3}{-3}=\frac{-8}{8}=-1✓$

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Summary