Vector Algebra — Class 12 IMO

Vectors and Components

A vector has both magnitude and direction (e.g. displacement, force), unlike a scalar which has magnitude only. In space a vector is written $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$.

Magnitude and unit vectors

The magnitude is $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}$. A unit vector in the direction of $\vec a$ is $\hat a=\dfrac{\vec a}{|\vec a|}$.

Types of vectors

Zero vector $\vec 0$: magnitude $0$, no definite direction.
Unit vector: magnitude $1$.
Equal vectors: same magnitude and direction.
Collinear/parallel: $\vec a=\lambda\vec b$ for some scalar $\lambda$.

Addition and scalar multiplication

Add componentwise; scalar $\lambda$ multiplies each component (triangle/parallelogram law geometrically). The position vector of a point $P(x,y,z)$ is $x\hat i+y\hat j+z\hat k$, and the vector from $A$ to $B$ is $\overrightarrow{AB}=\vec b-\vec a$ (position vector of $B$ minus that of $A$). The section formula gives the point dividing $AB$ in ratio $m:n$ as $\dfrac{m\vec b+n\vec a}{m+n}$.

Example 1: Find $|\vec a|$ for $\vec a=3\hat i+4\hat k$.

$|\vec a|=\sqrt{3^2+0^2+4^2}=\sqrt{25}=5.$

Example 2: Find the unit vector along $\vec a=2\hat i+3\hat j+6\hat k$.

$|\vec a|=\sqrt{4+9+36}=\sqrt{49}=7$. So $\hat a=\dfrac{1}{7}(2\hat i+3\hat j+6\hat k).$

Example 3: Find $\overrightarrow{AB}$ where $A(1,2,3)$, $B(4,6,3)$.

$\overrightarrow{AB}=\vec b-\vec a=(4-1)\hat i+(6-2)\hat j+(3-3)\hat k=3\hat i+4\hat j.$

Example 4: Find the midpoint of $A(2,0,4)$ and $B(0,2,0)$ as a position vector.

Midpoint $=\dfrac{\vec a+\vec b}{2}=\dfrac{(2\hat i+4\hat k)+(2\hat j)}{2}=\hat i+\hat j+2\hat k.$

Quick recap

$\vec a=a_1\hat i+a_2\hat j+a_3\hat k$; $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}$.
Unit vector $\hat a=\dfrac{\vec a}{|\vec a|}$.
$\overrightarrow{AB}=\vec b-\vec a$; section formula $\dfrac{m\vec b+n\vec a}{m+n}$.
Parallel vectors satisfy $\vec a=\lambda\vec b$.

✓ Quick check

If a = i + j and b = i − j, then a·b equals:

1×1 + 1×(−1) = 0.

The angle between two equal non-zero vectors is:

Equal vectors have the same direction.

Dot (Scalar) Product

The dot product of two vectors produces a scalar:

$$\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta = a_1b_1+a_2b_2+a_3b_3,$$

where $\theta$ is the angle between them.

Key consequences

Angle: $\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}$.
Perpendicularity: $\vec a\perp\vec b \iff \vec a\cdot\vec b=0$.
$\vec a\cdot\vec a=|\vec a|^2$.
Projection of $\vec a$ on $\vec b$: $\dfrac{\vec a\cdot\vec b}{|\vec b|}$.

The dot product is commutative ($\vec a\cdot\vec b=\vec b\cdot\vec a$) and distributive over addition.

Example 1: Find $\vec a\cdot\vec b$ for $\vec a=\hat i+2\hat j+3\hat k$, $\vec b=2\hat i-\hat j+\hat k$.

$=1\cdot2+2\cdot(-1)+3\cdot1=2-2+3=3.$

Example 2: Are $\vec a=\hat i+\hat j$ and $\vec b=\hat i-\hat j$ perpendicular?

$\vec a\cdot\vec b=1\cdot1+1\cdot(-1)=0$. Since the dot product is $0$, yes, they are perpendicular.

Example 3: Find the angle between $\vec a=\hat i$ and $\vec b=\hat i+\hat j$.

$\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\dfrac{1}{1\cdot\sqrt2}=\dfrac{1}{\sqrt2}$, so $\theta=45^\circ.$

Example 4: Find the projection of $\vec a=2\hat i+3\hat j$ on $\vec b=\hat i$.

Projection $=\dfrac{\vec a\cdot\vec b}{|\vec b|}=\dfrac{2}{1}=2.$

Quick recap

$\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=a_1b_1+a_2b_2+a_3b_3$ (a scalar).
$\vec a\perp\vec b \iff \vec a\cdot\vec b=0$; $\ \vec a\cdot\vec a=|\vec a|^2$.
$\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}$.
Projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$.

✓ Quick check

The value of i·i is:

A unit vector dotted with itself equals 1.

If θ is the angle between any two vectors a and b, then |a.b| = |a × b| when θ is equal to:

|a||b||cosθ| = |a||b||sinθ|. This implies |tanθ| = 1. The angle in the first quadrant is π/4 (45°).

Cross (Vector) Product

The cross product of two vectors produces a vector perpendicular to both:

$$\vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n = \begin{vmatrix}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix},$$

where $\hat n$ is the unit vector perpendicular to the plane of $\vec a,\vec b$ (right-hand rule).

Key consequences

Anti-commutative: $\vec a\times\vec b=-(\vec b\times\vec a)$.
Parallel test: $\vec a\parallel\vec b \iff \vec a\times\vec b=\vec 0$.
$|\vec a\times\vec b|$ is the area of the parallelogram with sides $\vec a,\vec b$; half of it is the triangle's area.
$\sin\theta=\dfrac{|\vec a\times\vec b|}{|\vec a||\vec b|}$.

Example 1: Find $\vec a\times\vec b$ for $\vec a=\hat i+\hat j$, $\vec b=\hat j+\hat k$.

$\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&0\\0&1&1\end{vmatrix}=\hat i(1\cdot1-0\cdot1)-\hat j(1\cdot1-0\cdot0)+\hat k(1\cdot1-1\cdot0)=\hat i-\hat j+\hat k.$

Example 2: Find the area of the parallelogram with adjacent sides $\vec a=\hat i+\hat j$ and $\vec b=\hat j+\hat k$.

From Example 1, $\vec a\times\vec b=\hat i-\hat j+\hat k$, so area $=|\vec a\times\vec b|=\sqrt{1+1+1}=\sqrt3$ square units.

Example 3: Are $\vec a=2\hat i+4\hat j$ and $\vec b=\hat i+2\hat j$ parallel?

$\vec a=2\vec b$, so they are parallel. Check: $\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\2&4&0\\1&2&0\end{vmatrix}=\vec 0$, confirming parallelism.

Example 4: Find the area of the triangle with sides $\vec a=\hat i+\hat j$, $\vec b=\hat j+\hat k$.

Area $=\tfrac12|\vec a\times\vec b|=\tfrac12\sqrt3=\dfrac{\sqrt3}{2}$ square units.

Quick recap

$\vec a\times\vec b$ is a vector $\perp$ both, magnitude $|\vec a||\vec b|\sin\theta$.
Anti-commutative: $\vec a\times\vec b=-\vec b\times\vec a$.
$\vec a\parallel\vec b \iff \vec a\times\vec b=\vec 0$.
$|\vec a\times\vec b|=$ area of parallelogram; triangle area $=\tfrac12|\vec a\times\vec b|$.

✓ Quick check

Riya walks 6 km east and then 8 km north. Her displacement magnitude is:

√(6²+8²)=10 km.

A shop board is hung using two strings. The tension in the strings are T₁ = 2î + 3ĵ and T₂ = −2î + 4ĵ. The total upward pull supporting the weight of the board is:

Total upward force = T₁ + T₂ = (2î + 3ĵ) + (−2î + 4ĵ) = 0î + 7ĵ = 7ĵ.