Vector Algebra
Complete JEE preparation for vectors — basics, dot and cross products, scalar and vector triple products, and geometric applications with competitive-level shortcuts.
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01
Vector Basics
Types of vectors, position vectors, components, unit vectors, and section formula — the foundation for all vector operations.
Types, Components & Unit Vectors
Definition — Vector
A vector is a quantity with both magnitude and direction. In 3D, $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ with magnitude $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.
Types of vectors:
- Zero vector $\vec{0}$: magnitude 0, arbitrary direction.
- Unit vector: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$, magnitude 1.
- Co-initial: same starting point. Collinear: parallel (or anti-parallel). Coplanar: lie in one plane.
- Position vector: $\vec{OP}$ of point $P(x,y,z)$ is $x\hat{i}+y\hat{j}+z\hat{k}$.
Section Formula
If $P$ divides $AB$ internally in ratio $m:n$: $\vec{OP} = \frac{m\vec{b}+n\vec{a}}{m+n}$.
Externally: $\vec{OP} = \frac{m\vec{b}-n\vec{a}}{m-n}$.
Midpoint: $\vec{OP} = \frac{\vec{a}+\vec{b}}{2}$. Centroid of triangle: $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$.
Externally: $\vec{OP} = \frac{m\vec{b}-n\vec{a}}{m-n}$.
Midpoint: $\vec{OP} = \frac{\vec{a}+\vec{b}}{2}$. Centroid of triangle: $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$.
JEE Trick — Collinearity
Three points $A, B, C$ are collinear if $\vec{AB} = \lambda\,\vec{AC}$ for some scalar $\lambda$. Equivalently, $\vec{AB} \times \vec{AC} = \vec{0}$.
📝 Example
Find the unit vector in the direction of $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.
$|\vec{a}| = \sqrt{4+9+36} = 7$. Unit vector $= \frac{1}{7}(2\hat{i}-3\hat{j}+6\hat{k})$.
02
Dot Product (Scalar Product)
Projection, angle between vectors, work done — dot product is the workhorse of JEE vector problems.
Dot Product & Properties
Definition — Dot Product
$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Key properties:
- $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ (commutative).
- $\vec{a} \cdot (\vec{b}+\vec{c}) = \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}$ (distributive).
- $\vec{a} \cdot \vec{a} = |\vec{a}|^2$.
- $\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1$; $\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0$.
Projection & Component
Projection of $\vec{a}$ on $\vec{b}$: $\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ (scalar).
Vector projection: $\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\;\vec{b}$.
JEE Trick: $|\vec{a}+\vec{b}|^2 = |\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2$ — expand like algebra using $\vec{a}\cdot\vec{a}=|\vec{a}|^2$.
Vector projection: $\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\;\vec{b}$.
JEE Trick: $|\vec{a}+\vec{b}|^2 = |\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2$ — expand like algebra using $\vec{a}\cdot\vec{a}=|\vec{a}|^2$.
Angle Between Vectors
$\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$.
Perpendicular: $\vec{a}\cdot\vec{b}=0$. Parallel (same direction): $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|$.
Perpendicular: $\vec{a}\cdot\vec{b}=0$. Parallel (same direction): $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|$.
📝 Example
If $|\vec{a}|=3$, $|\vec{b}|=4$, and $\vec{a}\cdot\vec{b}=6$, find the angle between them and $|\vec{a}-\vec{b}|$.
$\cos\theta = \frac{6}{3\cdot4} = \frac{1}{2}$, so $\theta = 60°$.
$|\vec{a}-\vec{b}|^2 = 9+16-2(6) = 13$, so $|\vec{a}-\vec{b}| = \sqrt{13}$.
$|\vec{a}-\vec{b}|^2 = 9+16-2(6) = 13$, so $|\vec{a}-\vec{b}| = \sqrt{13}$.
03
Cross Product (Vector Product)
Area of parallelogram, torque, and perpendicular vectors — cross product is key for JEE 3D and physics problems.
Cross Product & Properties
Definition — Cross Product
$\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta\;\hat{n}$, where $\hat{n}$ is the unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ (right-hand rule).
In component form: $$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix} = (a_2b_3-a_3b_2)\hat{i}-(a_1b_3-a_3b_1)\hat{j}+(a_1b_2-a_2b_1)\hat{k}$$
In component form: $$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix} = (a_2b_3-a_3b_2)\hat{i}-(a_1b_3-a_3b_1)\hat{j}+(a_1b_2-a_2b_1)\hat{k}$$
Key properties:
- $\vec{a}\times\vec{b} = -(\vec{b}\times\vec{a})$ (anti-commutative).
- $\vec{a}\times\vec{a} = \vec{0}$.
- $\hat{i}\times\hat{j}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, $\hat{k}\times\hat{i}=\hat{j}$ (cyclic).
- $|\vec{a}\times\vec{b}|$ = area of parallelogram with sides $\vec{a}$ and $\vec{b}$.
Area Formulas
Area of parallelogram: $|\vec{a}\times\vec{b}|$.
Area of triangle with sides $\vec{a}, \vec{b}$: $\frac{1}{2}|\vec{a}\times\vec{b}|$.
Area of triangle with vertices $A, B, C$: $\frac{1}{2}|\vec{AB}\times\vec{AC}|$.
JEE Shortcut: For diagonals $\vec{d}_1, \vec{d}_2$ of a parallelogram, area $= \frac{1}{2}|\vec{d}_1\times\vec{d}_2|$.
Area of triangle with sides $\vec{a}, \vec{b}$: $\frac{1}{2}|\vec{a}\times\vec{b}|$.
Area of triangle with vertices $A, B, C$: $\frac{1}{2}|\vec{AB}\times\vec{AC}|$.
JEE Shortcut: For diagonals $\vec{d}_1, \vec{d}_2$ of a parallelogram, area $= \frac{1}{2}|\vec{d}_1\times\vec{d}_2|$.
Lagrange's Identity
$|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2$.
JEE Trick: Use this when you know $|\vec{a}|$, $|\vec{b}|$, and $\vec{a}\cdot\vec{b}$ but need $|\vec{a}\times\vec{b}|$.
JEE Trick: Use this when you know $|\vec{a}|$, $|\vec{b}|$, and $\vec{a}\cdot\vec{b}$ but need $|\vec{a}\times\vec{b}|$.
📝 Example
Find the area of the triangle with vertices $A(1,1,2)$, $B(2,3,5)$, $C(1,5,5)$.
$\vec{AB}=(1,2,3)$, $\vec{AC}=(0,4,3)$. $\vec{AB}\times\vec{AC} = (6-12,\;0-3,\;4-0) = (-6,-3,4)$.
$|\vec{AB}\times\vec{AC}| = \sqrt{36+9+16}=\sqrt{61}$. Area $= \frac{\sqrt{61}}{2}$.
$|\vec{AB}\times\vec{AC}| = \sqrt{36+9+16}=\sqrt{61}$. Area $= \frac{\sqrt{61}}{2}$.
04
Scalar Triple Product
Volume of parallelepiped, coplanarity test, and JEE Advanced problems involving the box product $[\vec{a}\;\vec{b}\;\vec{c}]$.
Scalar Triple Product & Applications
Definition — Scalar Triple Product
$[\vec{a}\;\vec{b}\;\vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = \begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}$
This gives the signed volume of the parallelepiped formed by $\vec{a}, \vec{b}, \vec{c}$.
This gives the signed volume of the parallelepiped formed by $\vec{a}, \vec{b}, \vec{c}$.
Key properties:
- $[\vec{a}\;\vec{b}\;\vec{c}] = [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{c}\;\vec{a}\;\vec{b}]$ (cyclic permutation preserves value).
- Swapping two adjacent vectors changes the sign: $[\vec{a}\;\vec{b}\;\vec{c}] = -[\vec{b}\;\vec{a}\;\vec{c}]$.
- $[\vec{a}\;\vec{b}\;\vec{c}] = 0 \iff \vec{a}, \vec{b}, \vec{c}$ are coplanar.
Volume Formulas
Volume of parallelepiped: $|[\vec{a}\;\vec{b}\;\vec{c}]|$.
Volume of tetrahedron: $\frac{1}{6}|[\vec{a}\;\vec{b}\;\vec{c}]|$ (where $\vec{a}, \vec{b}, \vec{c}$ are edges from one vertex).
JEE Trick: Volume of tetrahedron with vertices $A, B, C, D$ is $\frac{1}{6}|[\vec{AB}\;\vec{AC}\;\vec{AD}]|$.
Volume of tetrahedron: $\frac{1}{6}|[\vec{a}\;\vec{b}\;\vec{c}]|$ (where $\vec{a}, \vec{b}, \vec{c}$ are edges from one vertex).
JEE Trick: Volume of tetrahedron with vertices $A, B, C, D$ is $\frac{1}{6}|[\vec{AB}\;\vec{AC}\;\vec{AD}]|$.
Coplanarity Test
Four points $A, B, C, D$ are coplanar $\iff [\vec{AB}\;\vec{AC}\;\vec{AD}] = 0$.
Three vectors are coplanar $\iff$ their scalar triple product is zero $\iff$ the $3\times3$ determinant is zero.
Three vectors are coplanar $\iff$ their scalar triple product is zero $\iff$ the $3\times3$ determinant is zero.
📝 Example
Find the volume of the tetrahedron with vertices $A(1,1,1)$, $B(2,1,3)$, $C(3,2,2)$, $D(3,3,4)$.
$\vec{AB}=(1,0,2)$, $\vec{AC}=(2,1,1)$, $\vec{AD}=(2,2,3)$.
$[\vec{AB}\;\vec{AC}\;\vec{AD}] = \begin{vmatrix}1&0&2\\2&1&1\\2&2&3\end{vmatrix} = 1(3-2)-0(6-2)+2(4-2) = 1+4 = 5$.
Volume $= \frac{|5|}{6} = \boxed{\frac{5}{6}}$.
$[\vec{AB}\;\vec{AC}\;\vec{AD}] = \begin{vmatrix}1&0&2\\2&1&1\\2&2&3\end{vmatrix} = 1(3-2)-0(6-2)+2(4-2) = 1+4 = 5$.
Volume $= \frac{|5|}{6} = \boxed{\frac{5}{6}}$.
05
Vector Triple Product & Applications in Geometry
The BAC-CAB rule, applications to lines, planes, and geometric proofs — Advanced-level vector techniques.
Vector Triple Product
Vector Triple Product (BAC-CAB Rule)
$$\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}$$
$$(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{c})\vec{a}$$
JEE Trick: Remember "BAC-CAB" — the outer vector dots with the far vector, times the near vector, minus the outer dots near, times far.
Important Identity
$\vec{a}\times(\vec{b}\times\vec{c}) + \vec{b}\times(\vec{c}\times\vec{a}) + \vec{c}\times(\vec{a}\times\vec{b}) = \vec{0}$ (Jacobi identity).
Geometric Applications
Applications Summary
Angle bisector: Unit vectors along the bisector of angle between $\vec{a}$ and $\vec{b}$: $\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}$ (internal) and $\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$ (external).
Foot of perpendicular from $P$ to line $\vec{r}=\vec{a}+\lambda\vec{b}$: $\lambda = \frac{(\vec{OP}-\vec{a})\cdot\vec{b}}{|\vec{b}|^2}$.
Projection of point on plane: Use $\vec{r} = \vec{p} + t\,\vec{n}$ and substitute in the plane equation.
Foot of perpendicular from $P$ to line $\vec{r}=\vec{a}+\lambda\vec{b}$: $\lambda = \frac{(\vec{OP}-\vec{a})\cdot\vec{b}}{|\vec{b}|^2}$.
Projection of point on plane: Use $\vec{r} = \vec{p} + t\,\vec{n}$ and substitute in the plane equation.
JEE Shortcut — Reciprocal System of Vectors
If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, the reciprocal vectors are:
$\vec{a}' = \frac{\vec{b}\times\vec{c}}{[\vec{a}\;\vec{b}\;\vec{c}]}$, $\vec{b}' = \frac{\vec{c}\times\vec{a}}{[\vec{a}\;\vec{b}\;\vec{c}]}$, $\vec{c}' = \frac{\vec{a}\times\vec{b}}{[\vec{a}\;\vec{b}\;\vec{c}]}$.
Key: $\vec{a}\cdot\vec{a}'=1$, $\vec{a}\cdot\vec{b}'=0$, etc.
Key: $\vec{a}\cdot\vec{a}'=1$, $\vec{a}\cdot\vec{b}'=0$, etc.
📝 Example
If $\vec{a}\times\vec{b}=\vec{c}\times\vec{d}$ and $\vec{a}\times\vec{c}=\vec{b}\times\vec{d}$, show that $\vec{a}-\vec{d}$ is parallel to $\vec{b}-\vec{c}$.
$\vec{a}\times\vec{b}-\vec{a}\times\vec{c} = \vec{c}\times\vec{d}-\vec{b}\times\vec{d}$.
$\vec{a}\times(\vec{b}-\vec{c}) = (\vec{c}-\vec{b})\times\vec{d} = -(\vec{b}-\vec{c})\times\vec{d} = \vec{d}\times(\vec{b}-\vec{c})$.
$(\vec{a}-\vec{d})\times(\vec{b}-\vec{c}) = \vec{0}$, so $\vec{a}-\vec{d}$ is parallel to $\vec{b}-\vec{c}$.
$\vec{a}\times(\vec{b}-\vec{c}) = (\vec{c}-\vec{b})\times\vec{d} = -(\vec{b}-\vec{c})\times\vec{d} = \vec{d}\times(\vec{b}-\vec{c})$.
$(\vec{a}-\vec{d})\times(\vec{b}-\vec{c}) = \vec{0}$, so $\vec{a}-\vec{d}$ is parallel to $\vec{b}-\vec{c}$.
★ Key Takeaways
- $\vec{a}\cdot\vec{b}=0$ means perpendicular; $\vec{a}\times\vec{b}=\vec{0}$ means parallel — never confuse the two conditions.
- Lagrange's identity $|\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2$ avoids computing cross products when magnitudes and dot product are known.
- Scalar triple product zero $\Leftrightarrow$ coplanar — this is the single most important coplanarity test in JEE.
- BAC-CAB rule: memorize $\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$ — it appears in JEE Advanced nearly every year.
- Volume of tetrahedron $= \frac{1}{6}$ of parallelepiped volume — do not forget the $\frac{1}{6}$ factor.
✎ Practice Problems
Problem 1
If $|\vec{a}|=2$, $|\vec{b}|=3$, and the angle between them is $60°$, find $|\vec{a}\times\vec{b}|$ and $\vec{a}\cdot\vec{b}$.
Show Solution ▼
$\vec{a}\cdot\vec{b}=2\cdot3\cdot\cos60°=3$. $|\vec{a}\times\vec{b}|=2\cdot3\cdot\sin60°=3\sqrt{3}$.
Problem 2
Show that $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=\hat{i}-\hat{j}-4\hat{k}$ are coplanar.
Show Solution ▼
$[\vec{a}\;\vec{b}\;\vec{c}]=\begin{vmatrix}1&2&3\\2&1&-1\\1&-1&-4\end{vmatrix}=1(-4-1)-2(-8+1)+3(-2-1)=(-5)+(14)+(-9)=0$. Hence coplanar.
Problem 3
Find a vector perpendicular to both $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, having magnitude $\sqrt{14}$.
Show Solution ▼
$\vec{a}\times\vec{b}=(2-3)\hat{i}-(−1-6)\hat{j}+(1+4)\hat{k}=-\hat{i}+7\hat{j}+5\hat{k}$. $|\vec{a}\times\vec{b}|=\sqrt{1+49+25}=\sqrt{75}=5\sqrt{3}$. Required vector $= \pm\frac{\sqrt{14}}{5\sqrt{3}}(-\hat{i}+7\hat{j}+5\hat{k})$.
Problem 4
If $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=3$, $|\vec{b}|=5$, $|\vec{c}|=7$, find the angle between $\vec{a}$ and $\vec{b}$.
Show Solution ▼
$\vec{c}=-(\vec{a}+\vec{b})$. $|\vec{c}|^2=|\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2$. $49=9+2\vec{a}\cdot\vec{b}+25$. $\vec{a}\cdot\vec{b}=\frac{15}{2}$. $\cos\theta=\frac{15/2}{3\cdot5}=\frac{1}{2}$. $\theta=60°$.
Problem 5
Evaluate $\vec{a}\times(\vec{b}\times\vec{c})$ if $\vec{a}=\hat{i}+\hat{j}$, $\vec{b}=\hat{j}+\hat{k}$, $\vec{c}=\hat{k}+\hat{i}$.
Show Solution ▼
Using BAC-CAB: $\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$. $\vec{a}\cdot\vec{c}=(1)(1)+(1)(0)+(0)(1)=1$. $\vec{a}\cdot\vec{b}=0+1+0=1$. Result $= 1(\hat{j}+\hat{k})-1(\hat{k}+\hat{i})=\hat{j}-\hat{i}=-\hat{i}+\hat{j}$.
🎯 Interactive MCQs
1. If $\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}$, then it is necessary that:
A $\vec{b}=\vec{c}$
B $\vec{a}\perp(\vec{b}-\vec{c})$ or $\vec{a}=\vec{0}$ or $\vec{b}=\vec{c}$
C $\vec{a}=\vec{0}$
D $\vec{a}\perp\vec{b}$
2. The area of the parallelogram with diagonals $\vec{d}_1 = 3\hat{i}+\hat{j}-2\hat{k}$ and $\vec{d}_2 = \hat{i}-3\hat{j}+4\hat{k}$ is:
A $10\sqrt{3}$
B $5\sqrt{3}$
C $\sqrt{300}$
D $15$
3. If $[\vec{a}\;\vec{b}\;\vec{c}]=6$, then $[\vec{a}+\vec{b}\;\;\vec{b}+\vec{c}\;\;\vec{c}+\vec{a}]$ equals:
A $6$
B $12$
C $18$
D $24$
4. If $|\vec{a}|=|\vec{b}|=1$ and $|\vec{a}+\vec{b}|=\sqrt{3}$, then $|\vec{a}-\vec{b}|$ is:
A $1$
B $\sqrt{2}$
C $\sqrt{3}$
D $2$
5. If $\vec{a}\times\vec{b}=\vec{b}\times\vec{c}\neq\vec{0}$, then $\vec{a}+\vec{c}$ equals:
A $\vec{0}$
B $2\vec{a}$
C $\lambda\vec{b}$ for some scalar $\lambda$
D A vector perpendicular to $\vec{b}$