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Class 12 Maths Notes

Vector Algebra Class 12 Notes

Complete, exam-ready notes on vector algebra: components and position vectors, the dot and cross products, projection of vectors and their geometric meaning — written for CBSE boards and JEE revision.

Class12SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is this chapter about in one line?

Vector algebra describes quantities with both magnitude and direction, and the dot and cross products let you measure angles, projections and areas.

Vectors and Components

Position vector

A vector has magnitude and direction. The position vector of point P(x,y,z)P(x,y,z) is r⃗=x i^+y j^+z k^\vec{r} = x\,\hat{i} + y\,\hat{j} + z\,\hat{k} with magnitude ∣r⃗∣=x2+y2+z2|\vec{r}| = \sqrt{x^2 + y^2 + z^2}.

AB⃗=b⃗−a⃗,a^=a⃗∣a⃗∣\vec{AB} = \vec{b} - \vec{a},\qquad \hat{a} = \frac{\vec{a}}{|\vec{a}|}
Vector between two points and unit vector

Dot (Scalar) Product

Dot product

The dot product is a scalar measuring how much two vectors point together: a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta. Two non-zero vectors are perpendicular exactly when their dot product is zero.

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣,projb⃗a⃗=a⃗⋅b⃗∣b⃗∣\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|},\qquad \text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}
Angle between vectors and projection

Cross (Vector) Product

Cross product

The cross product of two non-parallel vectors is a vector perpendicular to both, with magnitude ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta equal to the parallelogram area. For parallel vectors it is the zero vector.

a⃗×b⃗=∣i^j^k^a1a2a3b1b2b3∣,∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ\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},\qquad |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta
Cross product and its magnitude

Collinearity and Coplanarity

  • Points A, B, C are collinear if
  • AB⃗×AC⃗=0⃗\vec{AB} \times \vec{AC} = \vec{0}
  • (the displacement vectors are parallel).
  • Vectors are coplanar if their scalar triple product is zero:
  • a⃗⋅(b⃗×c⃗)=0\vec{a}\cdot(\vec{b}\times\vec{c}) = 0
  • .
  • Area of a parallelogram with sides
  • a⃗, b⃗\vec{a},\ \vec{b}
  • is
  • ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|
  • ; a triangle takes half of it.

Solved Examples

Example: Find the angle between a⃗=i^+2j^+2k^\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k} and b⃗=3i^+0j^+4k^\vec{b} = 3\hat{i} + 0\hat{j} + 4\hat{k}.

Solution: a⃗⋅b⃗=3+0+8=11\vec{a}\cdot\vec{b} = 3 + 0 + 8 = 11, ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=5|\vec{b}| = 5, so cos⁡θ=11/15\cos\theta = 11/15 and θ=cos⁡−1(11/15)\theta = \cos^{-1}(11/15).

Example: Find the area of the triangle with vertices (0,0,0), (1,2,3), (2,−1,1)(0,0,0),\ (1,2,3),\ (2,-1,1).

Solution: Take a⃗=i^+2j^+3k^\vec{a} = \hat{i}+2\hat{j}+3\hat{k} and b⃗=2i^−j^+k^\vec{b} = 2\hat{i}-\hat{j}+\hat{k}. Then a⃗×b⃗=(5,5,−5)\vec{a}\times\vec{b} = (5,5,-5) with length 535\sqrt{3}, and the triangle area is half of that: 532\frac{5\sqrt{3}}{2}.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Magnitude

∣a⃗∣=x2+y2+z2|\vec{a}| = \sqrt{x^2 + y^2 + z^2}

Dot product

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta

Cross product magnitude

∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta

Projection

projb⃗a⃗=a⃗⋅b⃗∣b⃗∣\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}

Parallelogram area

∣a⃗×b⃗∣|\vec{a}\times\vec{b}|

Coplanarity

a⃗⋅(b⃗×c⃗)=0\vec{a}\cdot(\vec{b}\times\vec{c}) = 0

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Dot product zero ⟺ perpendicular vectors; cross product zero ⟺ parallel vectors.
  • The cross product magnitude is the parallelogram area; the triangle area is half.
  • A unit vector in any direction is the vector divided by its magnitude.
  • Scalar triple product zero ⟺ the three vectors are coplanar.
  • Position vectors from the origin make displacement vectors by subtraction.

FAQ

Common questions

What is the difference between dot and cross product?

The dot product returns a scalar measuring how aligned two vectors are, while the cross product returns a new vector perpendicular to both, whose magnitude is the parallelogram area.

When is the dot product zero?

When the vectors are perpendicular (θ = 90°), since cos 90° = 0. Also if either vector is the zero vector.

How do you find a unit vector?

Divide the vector by its magnitude: â = a/|a|. The unit vector has the same direction but magnitude 1.

How is the area of a triangle related to vectors?

The area of a triangle with adjacent sides a and b is half the magnitude of their cross product: (1/2)|a × b|.

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