3.1 Q-16

Exercise 3.1 — Question 16

Problem Statement

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

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Key Concepts

Vector Between Two Points

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:

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

where $O$ is the origin.

Magnitude of a Vector

For a vector $\vec{v}=a\hat{i}+b\hat{j}+c\hat{k}$, the magnitude is:

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

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 one is a scalar multiple of the other, i.e., $\mathbf{a}=k\mathbf{b}$ for some scalar $k$. In component form:

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

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Worked Example

Given: $P(1,2,-1)$ and $Q(3,0,5)$

Step 1 — Find $\overrightarrow{PQ}$:

$\overrightarrow{PQ}=(3-1)\hat{i}+(0-2)\hat{j}+(5-(-1))\hat{k}=2\hat{i}-2\hat{j}+6\hat{k}$

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

$|\overrightarrow{PQ}|=\sqrt{(2{)}^2+(-2{)}^2+(6{)}^2}=\sqrt{4+4+36}=\sqrt{44}=2\sqrt{11}$

Step 3 — Unit vector along $\overrightarrow{PQ}$:

$\hat{u}=\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{2\hat{i}-2\hat{j}+6\hat{k}}{2\sqrt{11}}=\frac{1}{\sqrt{11}}\hat{i}-\frac{1}{\sqrt{11}}\hat{j}+\frac{3}{\sqrt{11}}\hat{k}$

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