Maths · JEE Main & Advanced 2026 · Vector Algebra

Vector Algebra notes, formulas & PYQs for JEE 2026

Complete chapter — position vector, types of vectors, addition & section formula, dot product, cross product, scalar triple product, vector triple product, projection of vectors, direction cosines & ratios, linear independence, 100+ worked examples, and JEE Main/Advanced PYQs with step-by-step solutions.

SECTIONS
10
SOLVED EXAMPLES
100+
READ TIME
~5h
COST
₹0

Table of Contents

Jump to any section
Vector Algebra JEE Notes, Formulas, PYQs
Vector Algebra — JEE Notes, Formulas, PYQs
1

Basics & Types of Vectors

A vector is a quantity that has both magnitude and direction, while a scalar has only magnitude. Vectors are typically represented by directed line segments, denoted as \(\vec{a}\) or \(\vec{AB}\) (from point A to point B). The magnitude (or length) is denoted by \(|\vec{a}|\) or \(|\vec{AB}|\).

1.1 Position Vector

Important Concept — Position Vector

The position vector of a point \(P(x, y, z)\) in 3D space relative to the origin \(O(0, 0, 0)\) is given by:

\[\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}\]

Its magnitude is \(|\vec{OP}| = \sqrt{x^2 + y^2 + z^2}\).

1.2 Types of Vectors

  • Zero Vector (Null Vector): A vector with zero magnitude and undefined direction. Denoted as \(\vec{0}\). Example: \(\vec{AA}\).
  • Unit Vector: A vector of unit magnitude. Denoted by \(\hat{a}\) (read as 'a hat').
  • Equal Vectors: Two vectors are equal if they have the same magnitude and direction, regardless of their initial points.
  • Negative Vector: A vector having the same magnitude as the original vector but opposite direction.
  • Collinear Vectors: Two or more vectors parallel to the same line, irrespective of their magnitude and direction.
  • Co-initial Vectors: Vectors having the same starting point.
  • Co-planar Vectors: Vectors lying in the same plane.

1.3 Unit Vector Formula

Most Repeated PYQ — Unit Vector

The unit vector in the direction of a given vector \(\vec{a}\) is obtained by dividing the vector by its own magnitude:

\[\hat{a} = \frac{\vec{a}}{|\vec{a}|}\]

PYQ — JEE Main 2024 (Jan 31 Shift 1)

Question: Let \(\vec{a} = 2\hat{i} + \hat{j} - \hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + 2\hat{k}\). Find a unit vector parallel to \(\vec{a} + \vec{b}\).

Solution:

First, find \(\vec{a} + \vec{b}\):

\[\vec{a} + \vec{b} = (2+1)\hat{i} + (1-1)\hat{j} + (-1+2)\hat{k} = 3\hat{i} + 0\hat{j} + 1\hat{k}\]

Calculate magnitude:

\[|\vec{a} + \vec{b}| = \sqrt{3^2 + 0^2 + 1^2} = \sqrt{9 + 0 + 1} = \sqrt{10}\]

Now, the unit vector is:

\[\frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})\]

Answer: \(\frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})\)

2

Addition & Section Formula

Vector addition follows the triangle law or parallelogram law. The section formula is vital for finding a point that divides a line segment in a specific ratio.

2.1 Triangle Law of Addition

If two vectors are represented as two sides of a triangle in sequence, then their resultant is represented by the third side drawn from the tail of the first to the head of the second: \(\vec{AB} + \vec{BC} = \vec{AC}\).

2.2 Parallelogram Law of Addition

If two vectors acting at the same point are represented by the adjacent sides of a parallelogram, then the resultant is represented by the diagonal passing through that common point: \[|\vec{R}| = \sqrt{a^2 + b^2 + 2ab\cos\theta}\] where \(a = |\vec{a}|\), \(b = |\vec{b}|\), and \(\theta\) is the angle between them.

2.3 Section Formula (Internal & External)

Most Repeated PYQ — Section Formula

The position vector \(\vec{r}\) of a point \(P\) dividing the line segment joining \(A(\vec{a})\) and \(B(\vec{b})\) in the ratio \(m:n\) is:

Internally:

\[\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}\]

Externally:

\[\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}\]

2.4 Midpoint Formula

If \(P\) is the midpoint of \(AB\), it divides \(AB\) in ratio \(1:1\):

\[\vec{r} = \frac{\vec{a} + \vec{b}}{2}\]

PYQ — JEE Main 2023 (Apr 8 Shift 1)

Question: If the position vectors of points \(A\) and \(B\) are \(\hat{i} + 2\hat{j} - 3\hat{k}\) and \(2\hat{i} - \hat{j} + 4\hat{k}\) respectively, find the position vector of a point \(C\) that divides \(AB\) externally in the ratio \(2:1\).

Solution:

Using external section formula \(\vec{c} = \frac{m\vec{b} - n\vec{a}}{m - n}\) with \(m=2, n=1\):

\[\vec{c} = \frac{2(2\hat{i} - \hat{j} + 4\hat{k}) - 1(\hat{i} + 2\hat{j} - 3\hat{k})}{2 - 1}\]
\[\vec{c} = (4\hat{i} - 2\hat{j} + 8\hat{k}) - (\hat{i} + 2\hat{j} - 3\hat{k})\]
\[\vec{c} = 3\hat{i} - 4\hat{j} + 11\hat{k}\]
\[3\hat{i} - 4\hat{j} + 11\hat{k}\]

Answer: \(3\hat{i} - 4\hat{j} + 11\hat{k}\)

3

Dot Product (Scalar Product)

The dot product (or scalar product) of two vectors results in a scalar quantity. It is highly tested in JEE Mains and Advanced.

3.1 Definition & Formula

Most Repeated PYQ — Dot Product

The dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is defined as the product of their magnitudes and the cosine of the angle \(\theta\) between them:

\[\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\]

In component form, for \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\):

\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]

3.2 Properties of Dot Product

  • Commutative: \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\)
  • Distributive over addition: \(\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}\)
  • \(\vec{a} \cdot \vec{a} = |\vec{a}|^2\)
  • \(\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\)
  • \(\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0\)
  • Perpendicularity: Two non-zero vectors are perpendicular if and only if \(\vec{a} \cdot \vec{b} = 0\).
  • Cauchy-Schwartz Inequality: \(|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|\)

3.3 Angle Between Two Vectors

Important — Angle Formula

The angle \(\theta\) between vectors \(\vec{a}\) and \(\vec{b}\) is given by:

\[\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\]

PYQ — JEE Main 2024 (Apr 4 Shift 2)

Question: If \(\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}\), find the angle between \(\vec{a}\) and \(\vec{b}\).

Solution:

Find dot product: \(\vec{a} \cdot \vec{b} = (2)(1) + (1)(-2) + (2)(2) = 2 - 2 + 4 = 4\)

Find magnitudes:

\[|\vec{a}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3\]
\[|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3\]

Calculate cosine of angle:

\[\cos\theta = \frac{4}{3 \times 3} = \frac{4}{9}\]
\[\theta = \cos^{-1}\left(\frac{4}{9}\right)\]

Answer: \(\cos^{-1}\left(\frac{4}{9}\right)\)

4

Cross Product (Vector Product)

The cross product of two vectors results in a vector quantity. It is perpendicular to both vectors and represents the area of the parallelogram formed by them.

4.1 Definition & Formula

Most Repeated PYQ — Cross Product

The cross product of \(\vec{a}\) and \(\vec{b}\) is a vector defined as:

\[\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin\theta \, \hat{n}\]

where \(\hat{n}\) is the unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) (determined by the right-hand rule).

In determinant form:

\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}\]

4.2 Properties of Cross Product

  • Anti-commutative: \(\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\)
  • Distributive: \(\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}\)
  • Not associative: \(\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}\)
  • \(\vec{a} \times \vec{a} = \vec{0}\)
  • \(\hat{i} \times \hat{j} = \hat{k}, \hat{j} \times \hat{k} = \hat{i}, \hat{k} \times \hat{i} = \hat{j}\)
  • Parallel vectors: Two non-zero vectors are parallel iff \(\vec{a} \times \vec{b} = \vec{0}\).
  • Lagrange's Identity: \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2\)

4.3 Geometrical Interpretation (Area)

Most Repeated PYQ — Area of Triangle/Parallelogram

The magnitude of the cross product gives the area of the parallelogram formed by \(\vec{a}\) and \(\vec{b}\) as adjacent sides:

\[\text{Area of Parallelogram} = |\vec{a} \times \vec{b}|\]

The area of the triangle formed by \(\vec{a}\) and \(\vec{b}\) as adjacent sides is:

\[\text{Area of Triangle} = \frac{1}{2} |\vec{a} \times \vec{b}|\]

PYQ — JEE Main 2023 (Jan 25 Shift 2)

Question: Find the area of the triangle with vertices \(A(1, 1, 1)\), \(B(1, 2, 3)\), and \(C(2, 1, 2)\).

Solution:

Find vectors \(\vec{AB}\) and \(\vec{AC}\):

\[\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = \hat{j} + 2\hat{k}\]
\[\vec{AC} = (2-1)\hat{i} + (1-1)\hat{j} + (2-1)\hat{k} = \hat{i} + \hat{k}\]

Find \(\vec{AB} \times \vec{AC}\):

\[\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 2 \cdot 0) - \hat{j}(0 \cdot 1 - 2 \cdot 1) + \hat{k}(0 \cdot 0 - 1 \cdot 1)\]
\[= \hat{i}(1) - \hat{j}(-2) + \hat{k}(-1) = \hat{i} + 2\hat{j} - \hat{k}\]

Magnitude:

\[|\vec{AB} \times \vec{AC}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\]

Area of triangle = \(\frac{1}{2} |\vec{AB} \times \vec{AC}|\)

\[\text{Area} = \frac{\sqrt{6}}{2} \text{ sq. units}\]

Answer: \(\frac{\sqrt{6}}{2}\)

5

Scalar Triple Product

The scalar triple product of three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) is a scalar quantity, denoted by \([\vec{a} \, \vec{b} \, \vec{c}]\). It represents the volume of a parallelepiped.

5.1 Definition & Formula

Most Repeated PYQ — STP

The scalar triple product is defined as \(\vec{a} \cdot (\vec{b} \times \vec{c})\), evaluated as a determinant:

\[[\vec{a} \, \vec{b} \, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\]

5.2 Geometrical Meaning (Volume)

Most Repeated PYQ — Volume of Parallelepiped

The absolute value of the scalar triple product gives the volume of the parallelepiped formed by the three vectors as adjacent edges:

\[V = |[\vec{a} \, \vec{b} \, \vec{c}]|\]

Volume of a tetrahedron formed by \(\vec{a}, \vec{b}, \vec{c}\) is:

\[V_{tetra} = \frac{1}{6} |[\vec{a} \, \vec{b} \, \vec{c}]|\]

5.3 Properties of Scalar Triple Product

  • Cyclic permutation: \([\vec{a} \, \vec{b} \, \vec{c}] = [\vec{b} \, \vec{c} \, \vec{a}] = [\vec{c} \, \vec{a} \, \vec{b}]\)
  • Swapping adjacent vectors changes sign: \([\vec{a} \, \vec{b} \, \vec{c}] = -[\vec{b} \, \vec{a} \, \vec{c}]\)
  • If any two vectors are equal, \([\vec{a} \, \vec{a} \, \vec{c}] = 0\).
  • Coplanar vectors: Three vectors are coplanar if and only if their scalar triple product is zero: \([\vec{a} \, \vec{b} \, \vec{c}] = 0\).

PYQ — JEE Main 2022 (July 27 Shift 1)

Question: Find the volume of the parallelepiped whose adjacent edges are represented by the vectors \(\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}\), \(\vec{b} = \hat{i} + \hat{j} - \hat{k}\), and \(\vec{c} = 3\hat{i} + 2\hat{j} - 2\hat{k}\).

Solution:

Evaluate the determinant \([\vec{a} \, \vec{b} \, \vec{c}]\):

\[[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} 2 & -3 & 1 \\ 1 & 1 & -1 \\ 3 & 2 & -2 \end{vmatrix}\]

Expand along first row:

\[= 2((1)(-2) - (-1)(2)) - (-3)((1)(-2) - (-1)(3)) + 1((1)(2) - (1)(3))\]
\[= 2(-2 + 2) + 3(-2 + 3) + 1(2 - 3)\]
\[= 2(0) + 3(1) + 1(-1) = 0 + 3 - 1 = 2\]

Volume = \(|[\vec{a} \, \vec{b} \, \vec{c}]|\)

\[V = 2 \text{ cubic units}\]

Answer: 2

6

Vector Triple Product (VTP)

The vector triple product of three vectors \(\vec{a}, \vec{b}, \vec{c}\) is defined as \(\vec{a} \times (\vec{b} \times \vec{c})\). The result is a vector lying in the plane of \(\vec{b}\) and \(\vec{c}\).

6.1 Formula (BAC-CAB Rule)

Most Repeated PYQ — VTP Formula

The vector triple product is evaluated using the famous BAC-CAB rule:

\[\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\]

Similarly, for \((\vec{a} \times \vec{b}) \times \vec{c}\) (note the different bracket position!):

\[(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}\]

6.2 Properties & Coplanarity

  • The vector \(\vec{a} \times (\vec{b} \times \vec{c})\) is perpendicular to \(\vec{b} \times \vec{c}\), and hence lies in the plane of \(\vec{b}\) and \(\vec{c}\).
  • VTP is not associative: \(\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}\)
  • Jacobi Identity: \(\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = \vec{0}\)

PYQ — JEE Advanced 2023 (Paper 1)

Question: Let \(\vec{a} = \hat{i} + \hat{j} - \hat{k}\), \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), and \(\vec{c} = -\hat{i} + \hat{j} + \hat{k}\). Evaluate \(\vec{a} \times (\vec{b} \times \vec{c})\).

Solution:

Using formula \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\)

Calculate dot products:

\[\vec{a} \cdot \vec{c} = (1)(-1) + (1)(1) + (-1)(1) = -1 + 1 - 1 = -1\]
\[\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1\]

Substitute into the formula:

\[= (-1)\vec{b} - (-1)\vec{c} = -\vec{b} + \vec{c}\]
\[= -(\hat{i} - \hat{j} + \hat{k}) + (-\hat{i} + \hat{j} + \hat{k})\]
\[= -\hat{i} + \hat{j} - \hat{k} - \hat{i} + \hat{j} + \hat{k}\]
\[-2\hat{i} + 2\hat{j}\]

Answer: \(-2\hat{i} + 2\hat{j}\)

7

Projection of Vectors

The projection of a vector onto another gives the component of the first vector in the direction of the second. This is a vital concept frequently asked in JEE Mains.

7.1 Scalar Projection (Component)

Most Repeated PYQ — Projection Formula

The scalar projection (or component) of vector \(\vec{a}\) in the direction of vector \(\vec{b}\) is:

\[\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = |\vec{a}| \cos\theta\]

7.2 Vector Projection

The vector projection of \(\vec{a}\) on \(\vec{b}\) is the scalar projection multiplied by the unit vector in the direction of \(\vec{b}\):

\[\text{Vector projection of } \vec{a} \text{ on } \vec{b} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}\]

PYQ — JEE Main 2021 (Sep 1 Shift 1)

Question: Find the projection of the vector \(\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}\) on the vector \(\vec{b} = \hat{i} + 2\hat{j} + \hat{k}\).

Solution:

Find dot product: \(\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10\)

Find magnitude of \(\vec{b}\): \(|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}\)

Calculate projection:

\[\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{10}{\sqrt{6}}\]
\[\frac{10}{\sqrt{6}}\]

Answer: \(\frac{10}{\sqrt{6}}\)

8

Direction Cosines & Direction Ratios

The direction cosines of a vector are the cosines of the angles that the vector makes with the positive x, y, and z axes. Direction ratios are any three numbers proportional to the direction cosines.

8.1 Direction Cosines

Important — Direction Cosines

If a vector \(\vec{OP}\) makes angles \(\alpha, \beta, \gamma\) with the x, y, and z axes respectively, then its direction cosines (DCs) are:

\[l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma\]

The fundamental relation between direction cosines is:

\[l^2 + m^2 + n^2 = 1\]

8.2 Direction Ratios

If \(a, b, c\) are direction ratios (DRs) of a vector, then its direction cosines are:

\[l = \pm \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \pm \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \pm \frac{c}{\sqrt{a^2 + b^2 + c^2}}\]

For a position vector \(x\hat{i} + y\hat{j} + z\hat{k}\), the direction ratios are simply \(x, y, z\).

PYQ — JEE Main 2020 (Sep 4 Shift 1)

Question: A vector \(\vec{r}\) makes angles \(\frac{\pi}{3}\) with \(\hat{i}\) and \(\frac{\pi}{4}\) with \(\hat{j}\). Find the angle made by \(\vec{r}\) with \(\hat{k}\).

Solution:

Let the angle with \(\hat{k}\) be \(\gamma\). The direction cosines are:

\[l = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\]
\[m = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\]
\[n = \cos\gamma\]

Using \(l^2 + m^2 + n^2 = 1\):

\[\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2\gamma = 1\]
\[\frac{1}{4} + \frac{1}{2} + \cos^2\gamma = 1\]
\[\frac{3}{4} + \cos^2\gamma = 1 \implies \cos^2\gamma = \frac{1}{4}\]
\[\cos\gamma = \pm \frac{1}{2} \implies \gamma = \frac{\pi}{3} \text{ or } \frac{2\pi}{3}\]
\[\gamma = \frac{\pi}{3} \text{ or } \frac{2\pi}{3}\]

Answer: \(\frac{\pi}{3}\) or \(\frac{2\pi}{3}\)

9

Coplanar & Linearly Independent Vectors

Three vectors are coplanar if they lie in the same plane. A set of vectors is linearly independent if no vector in the set can be expressed as a linear combination of the others.

9.1 Condition for Coplanarity

Most Repeated PYQ — Coplanarity Condition

Three vectors \(\vec{a}, \vec{b}, \vec{c}\) are coplanar if and only if their scalar triple product is zero:

\[[\vec{a} \, \vec{b} \, \vec{c}] = 0\]

Alternatively, three vectors are coplanar if one can be expressed as a linear combination of the other two: \(\vec{c} = x\vec{a} + y\vec{b}\).

9.2 Linear Independence

A set of vectors \(\vec{a}_1, \vec{a}_2, \dots, \vec{a}_n\) is said to be linearly independent if the equation:

\[x_1\vec{a}_1 + x_2\vec{a}_2 + \dots + x_n\vec{a}_n = \vec{0}\]

holds only when \(x_1 = x_2 = \dots = x_n = 0\). Otherwise, they are linearly dependent.

For three 3D vectors, linear independence is equivalent to non-coplanarity, i.e., \([\vec{a} \, \vec{b} \, \vec{c}] \neq 0\).

PYQ — JEE Main 2019 (Apr 10 Shift 2)

Question: Find the value of \(\lambda\) for which the four points with position vectors \(6\hat{i} - 7\hat{j}\), \(16\hat{i} - 19\hat{j} - 4\hat{k}\), \(\lambda\hat{i} - 3\hat{j} + 6\hat{k}\), and \(2\hat{i} - 5\hat{j} + 10\hat{k}\) are coplanar.

Solution:

Let the points be \(A, B, C, D\). Find vectors \(\vec{AB}, \vec{AC}, \vec{AD}\):

\[\vec{AB} = 10\hat{i} - 12\hat{j} - 4\hat{k}\]
\[\vec{AC} = (\lambda - 6)\hat{i} + 4\hat{j} + 6\hat{k}\]
\[\vec{AD} = -4\hat{i} + 2\hat{j} + 10\hat{k}\]

For the points to be coplanar, \([\vec{AB} \, \vec{AC} \, \vec{AD}] = 0\):

\[\begin{vmatrix} 10 & -12 & -4 \\ \lambda - 6 & 4 & 6 \\ -4 & 2 & 10 \end{vmatrix} = 0\]

Expand the determinant:

\[10(40 - 12) - (-12)(10\lambda - 60 + 24) - 4(2\lambda - 12 + 16) = 0\]
\[10(28) + 12(10\lambda - 36) - 4(2\lambda + 4) = 0\]
\[280 + 120\lambda - 432 - 8\lambda - 16 = 0\]
\[112\lambda - 168 = 0 \implies \lambda = \frac{168}{112}\]
\[\lambda = \frac{3}{2}\]

Answer: \(\lambda = 1.5\)

10

Key Results & Identities

A quick summary of the most important formulas and identities that frequently appear in competitive exams.

Most Repeated PYQ — Formulas

1. Dot Product: \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3\)

2. Cross Product: \(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\)

3. Projection of \(\vec{a}\) on \(\vec{b}\): \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)

4. Vector Triple Product: \(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\)

5. Volume of Parallelepiped: \(|[\vec{a} \, \vec{b} \, \vec{c}]|\)

6. Area of Parallelogram: \(|\vec{a} \times \vec{b}|\)

7. Area of Triangle: \(\frac{1}{2}|\vec{a} \times \vec{b}|\)

8. Lagrange's Identity: \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2\)

9. Direction Cosines: \(l^2 + m^2 + n^2 = 1\)

10. Angle between vectors: \(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)

PYQ — JEE Main 2024 (Apr 6 Shift 1)

Question: If \(|\vec{a}| = 3, |\vec{b}| = 4\) and \(\vec{a} \cdot \vec{b} = 6\), find \(|\vec{a} \times \vec{b}|\).

Solution:

Using Lagrange's Identity: \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2\)

\[|\vec{a} \times \vec{b}|^2 = (3)^2(4)^2 - (6)^2\]
\[|\vec{a} \times \vec{b}|^2 = 9 \times 16 - 36 = 144 - 36 = 108\]
\[|\vec{a} \times \vec{b}| = \sqrt{108} = 6\sqrt{3}\]
\[6\sqrt{3}\]

Answer: \(6\sqrt{3}\)

Previous Year Questions Analysis

JEE Main (Last 10 Years)

  • ✓ Dot Product & Angle: 22%
  • ✓ Projection of Vectors: 15%
  • ✓ Cross Product (Area): 15%
  • ✓ Scalar Triple Product (Volume): 12%
  • ✓ Vector Triple Product: 10%
  • ✓ Section Formula / Position Vector: 8%
  • ✓ Unit Vector / Perpendicular Vector: 10%
  • ✓ Direction Cosines: 8%

JEE Advanced (Last 10 Years)

  • ✓ Vector Triple Product (BAC-CAB): 25%
  • ✓ Coplanarity / Linear Independence: 20%
  • ✓ Mixed problems (Lines + Vectors): 20%
  • ✓ Scalar Triple Product properties: 15%
  • ✓ Geometrical Applications: 20%

Top 15 Most Repeated Question Types

  1. Find angle between two vectors using \(\vec{a} \cdot \vec{b}\).
  2. Projection of \(\vec{a}\) on \(\vec{b}\) and vice versa.
  3. Area of triangle/parallelogram using cross product.
  4. Volume of parallelepiped/tetrahedron using STP.
  5. Evaluate Vector Triple Product using BAC-CAB rule.
  6. Find unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) (\(\vec{a} \times \vec{b}\)).
  7. Position vector dividing line segment internally/externally.
  8. Find \(\lambda\) for which points are coplanar (\([\vec{a} \, \vec{b} \, \vec{c}] = 0\)).
  9. Given \(|\vec{a}|, |\vec{b}|\), and \(\vec{a} \cdot \vec{b}\), find \(|\vec{a} \times \vec{b}|\) (Lagrange's identity).
  10. Verify linear independence of vectors.
  11. Find direction cosines given angles with axes.
  12. Compute dot/cross product from given components.
  13. Find a vector of given magnitude perpendicular to two vectors.
  14. Vector equation of a line passing through two points.
  15. Shortest distance between two skew lines (uses cross product).

Weightage Analysis

JEE Main: 8 marks (2 questions)
JEE Advanced: 4-8 marks (1-2 questions)
Difficulty Level: Easy to Moderate
Time Required: 4-5 hours practice

Practice Problem Set

Level 1: Basic (JEE Main Standard)

  1. Find the unit vector in the direction of \(2\hat{i} - 3\hat{j} + 6\hat{k}\).
  2. Find the dot product of \(\hat{i} + 2\hat{j} - 3\hat{k}\) and \(2\hat{i} - \hat{j} + \hat{k}\).
  3. Find the projection of \(\hat{i} + \hat{j}\) on \(\hat{i} - \hat{j}\).
  4. Find the area of the parallelogram formed by \(\hat{i} + \hat{j}\) and \(\hat{i} + 2\hat{j}\).
  5. If \([\vec{a} \, \vec{b} \, \vec{c}] = 5\), find \([\vec{b} \, \vec{a} \, \vec{c}]\).

Level 2: Intermediate (JEE Main/Advanced)

  1. Show that the vectors \(\hat{i} - 2\hat{j} + 3\hat{k}\), \(-2\hat{i} + 3\hat{j} - 4\hat{k}\), and \(\hat{i} - 3\hat{j} + 5\hat{k}\) are coplanar.
  2. Find the angle between the vectors \(2\hat{i} - \hat{j} + 2\hat{k}\) and \(3\hat{i} + 2\hat{j} - 6\hat{k}\).
  3. Find the volume of the parallelepiped formed by \(\hat{i} - \hat{j} + \hat{k}\), \(2\hat{i} + \hat{j} - \hat{k}\), and \(-\hat{i} + 2\hat{j} + 3\hat{k}\).
  4. Evaluate \(\vec{a} \times (\vec{b} \times \vec{c})\) if \(\vec{a} = \hat{i} + \hat{j}\), \(\vec{b} = \hat{j} + \hat{k}\), \(\vec{c} = \hat{k} + \hat{i}\).
  5. If \(|\vec{a}| = 5, |\vec{b}| = 3, |\vec{a} \times \vec{b}| = 12\), find \(\vec{a} \cdot \vec{b}\).

Level 3: Advanced (JEE Advanced/Olympiad)

  1. Prove that \(\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = \vec{0}\) (Jacobi Identity).
  2. If \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors such that \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), find the value of \(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}\).
  3. Find the value of \(\lambda\) such that the vectors \(\vec{a} = \hat{i} + \lambda\hat{j} + \hat{k}\), \(\vec{b} = \lambda\hat{i} + \hat{j} + \lambda\hat{k}\), and \(\vec{c} = \hat{i} + \hat{j} + \lambda\hat{k}\) are coplanar.
  4. Prove Lagrange's identity: \(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\).

> FAQ

Frequently Asked Questions

What is a unit vector?

A unit vector is a vector of unit magnitude (length 1). It is obtained by dividing a vector by its own magnitude. If a is a vector, its unit vector is â = a / |a|.

What is the formula for the dot product of two vectors?

The dot product (or scalar product) of two vectors a and b is given by a · b = |a| |b| cos θ, where θ is the angle between them. In component form, if a = a₁î + a₂ĵ + a₃k̂ and b = b₁î + b₂ĵ + b₃k̂, then a · b = a₁b₁ + a₂b₂ + a₃b₃.

What is the vector triple product?

The vector triple product of three vectors a, b, c is defined as a × (b × c). It is a vector quantity and is evaluated using the formula a × (b × c) = (a · c)b - (a · b)c.

What is the geometrical meaning of the scalar triple product?

The absolute value of the scalar triple product [a b c] = a · (b × c) represents the volume of the parallelepiped formed by the three vectors a, b, and c as adjacent edges. If it is zero, the vectors are coplanar.

How do you find the projection of vector a on vector b?

The scalar projection (or component) of vector a in the direction of vector b is given by (a · b) / |b|. The vector projection is ((a · b) / |b|²) b.

How many questions come from Vector Algebra in JEE Mains?

On average, 2 questions (8 marks) appear from Vector Algebra in every JEE Main session. Most repeated topics are dot product, cross product, projection of vectors, scalar triple product, and finding unit vectors.