Exercise 3.1 — Q3: Vectors in Space

Rectangular Coordinate System in Space

In 3D space, any point $P$ is located using an ordered triple $(x,y,z)$ with respect to three mutually perpendicular axes:

• $x$-axis — horizontal (left–right)
• $y$-axis — horizontal (forward–backward)
• $z$-axis — vertical (up–down)

The origin $O=(0,0,0)$ is the intersection of all three axes.

Unit Vectors and Components

Three standard unit vectors define the positive directions of the axes:

Any vector $\vec{v}$ in space can be written in component form: $\vec{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$ where $v_1$, $v_2$, $v_3$ are the scalar components along the $x$-, $y$-, and $z$-axes respectively.

Magnitude of a Vector

The magnitude (length) of $\vec{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$ is: $|\vec{v}|=\sqrt{v_1^2+v_2^2+v_3^2}$

This follows from the 3D extension of the Pythagorean theorem.

Vector Between Two Points

Given points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, the vector $\overrightarrow{PQ}$ is: $\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

Its magnitude is: $|\overrightarrow{PQ}|=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

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: $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}$ i.e., $\mathbf{a}=k\mathbf{b}$ for some scalar $k$.

Worked Example

Q3: Given $P(1,-2,3)$ and $Q(4,0,-1)$, find $\overrightarrow{PQ}$ and $|\overrightarrow{PQ}|$.

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

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