Is it correct to say: Confusing dot product (scalar) with cross product (vector)?

a · b is a NUMBER. a × b is a VECTOR. Writing '|a·b|' vs '|a×b|' loses marks if the wrong one is used.

Is it correct to say: Writing a × b = b × a?

Cross product is anti-commutative: a × b = −(b × a). Swapping flips the sign.

Is it correct to say: Computing 'unit vector perpendicular to a and b' as just a × b?

You must DIVIDE by |a × b| to get a unit vector: n̂ = (a × b)/|a × b|.

Is it correct to say: Using |a||b|cos θ for the cross product magnitude?

Cross product uses SIN: |a × b| = |a||b| sin θ. Dot product uses COS.

Is it correct to say: Forgetting the negative sign inside the j-component of the i, j, k determinant?

Cofactor sign at (1, 2) is negative. j-component of a × b is −(a₁b₃ − a₃b₁), i.e. a₃b₁ − a₁b₃. Miss the sign and the direction reverses.

Is it correct to say: Claiming three vectors are coplanar just because two are parallel?

Coplanarity means [a b c] = 0. Two vectors are always coplanar; the third one being in the same plane is the extra condition.

Vector Algebra

Vectors in 3-D, dot product, cross product, scalar triple product, projection, direction cosines — NCERT Class 12 Maths Ch 10

📐 14 formulas✏️ 3 examples🎯 6 practice⚖️ 6-8 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Cross product using the i, j, k determinant is asked every year — usually as part of a 4-mark question.
→Area of triangle = ½|a × b| for triangle formed by two adjacent side vectors — memorise this.
→Scalar triple product [a b c] = a · (b × c). If it is zero, the three vectors are COPLANAR.
→Direction cosines l, m, n always satisfy l² + m² + n² = 1 — use this as a check.
→For 'find unit vector perpendicular to both a and b' use (a × b) / |a × b|. Do NOT forget to divide by the magnitude.

📐 Formulas(14)

Vector in Component Form

a = a_{1} i^+ a_{2} j^+ a_{3} k^

Magnitude of a Vector★ Board fav

∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

Unit Vector

a^= \frac{a}{∣ a ∣}

Direction Cosines★ Board fav

l = cos α, m = cos β, n = cos γ; l^{2} + m^{2} + n^{2} = 1

Dot Product (Scalar Product)★ Board fav

a \cdot b = ∣ a ∣∣ b ∣ cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}

Angle Between Vectors★ Board fav

cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}

Projection of a on b

proj_{b} a = \frac{a \cdot b}{∣ b ∣}

Cross Product (Vector Product)★ Board fav

a \times b = i^a_{1} b_{1} j^a_{2} b_{2} k^a_{3} b_{3}

Magnitude of Cross Product★ Board fav

∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ

Area of Parallelogram / Triangle★ Board fav

Area_{parallelogram} = ∣ a \times b ∣, Area_{triangle} = \frac{1}{2} ∣ a \times b ∣

Unit Vector Perpendicular to Both

n^= \frac{a \times b}{∣ a \times b ∣}

Scalar Triple Product★ Board fav

[a b c] = a \cdot (b \times c) = a_{1} b_{1} c_{1} a_{2} b_{2} c_{2} a_{3} b_{3} c_{3}

Section Formula (Internal)

r = \frac{m b + n a}{m + n}

Non-commutativity of Cross Product

a \times b = - b \times a

✏️ Solved Examples

1Solved Exampleeasy2 steps

Find a unit vector in the direction of a = 3î − ĵ + 2k̂.

Compute the magnitude.

∣ a ∣ = 3^{2} + (- 1)^{2} + 2^{2} = 14

2Solved Exampleboard4 steps

Given a = 2î − ĵ + 2k̂ and b = î + 3ĵ − 2k̂, find the angle between them and the area of the parallelogram they form.

Compute the dot product and magnitudes.

a \cdot b = 2 (1) + (- 1) (3) + 2 (- 2) = - 5, ∣ a ∣ = 3, ∣ b ∣ = 14

3Solved ExampleHOTS5 steps

Show that vectors a = î + 2ĵ + 3k̂, b = 2î − ĵ + 4k̂, c = 3î + ĵ + 7k̂ are coplanar.

Vectors are coplanar iff [a b c] = 0. Set up the determinant.

[a b c] = 123 2 - 1 1 347

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Confusing dot product (scalar) with cross product (vector)
✓a · b is a NUMBER. a × b is a VECTOR. Writing '|a·b|' vs '|a×b|' loses marks if the wrong one is used.
2
Writing a × b = b × a
✓Cross product is anti-commutative: a × b = −(b × a). Swapping flips the sign.
3
Computing 'unit vector perpendicular to a and b' as just a × b
✓You must DIVIDE by |a × b| to get a unit vector: n̂ = (a × b)/|a × b|.
4
Using |a||b|cos θ for the cross product magnitude
✓Cross product uses SIN: |a × b| = |a||b| sin θ. Dot product uses COS.
5
Forgetting the negative sign inside the j-component of the i, j, k determinant
✓Cofactor sign at (1, 2) is negative. j-component of a × b is −(a₁b₃ − a₃b₁), i.e. a₃b₁ − a₁b₃. Miss the sign and the direction reverses.
6
Claiming three vectors are coplanar just because two are parallel
✓Coplanarity means [a b c] = 0. Two vectors are always coplanar; the third one being in the same plane is the extra condition.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the projection of a = î + 3ĵ + 7k̂ on b = 7î − ĵ + 8k̂.
Q2
Find a unit vector perpendicular to both a = 2î + ĵ + k̂ and b = î − ĵ + k̂.
Q3
The position vectors of A and B are î + 2ĵ − k̂ and 3î + 4ĵ + k̂. Find the position vector of the midpoint of AB.
Q4
Find the area of the triangle with vertices A(1, 1, 1), B(2, 3, 5), C(1, 5, 5).
Q5
Find the value of λ so that a = î + λĵ + k̂ and b = 2î + ĵ − k̂ are perpendicular.
Q6
Find the volume of the parallelepiped with edges a = î − 2ĵ + 3k̂, b = 2î + ĵ − k̂, c = ĵ + k̂.

📝 Notes

Vector Algebra

Chapter 10 gives the algebraic language of 3-D geometry. A vector has magnitude and direction; a scalar has only magnitude. The two products — dot and cross — encode angle and area/volume respectively.

Vector vs scalar operations

Dot product a · b: scalar. Measures how much of one vector lies along the other. Zero ⇔ perpendicular.
Cross product a × b: vector. Points perpendicular to both, magnitude equal to the parallelogram area they span. Zero ⇔ parallel.
Scalar triple product [a b c] = a · (b × c): scalar. Volume of parallelepiped. Zero ⇔ coplanar.

Right-hand rule

For a × b, curl the fingers of the right hand from a to b through the smaller angle; the thumb points along a × b. This is the ONLY way to get the direction right.

Choosing the correct product

Angle or projection — dot product.
Perpendicular direction, area of triangle/parallelogram, torque — cross product.
Coplanarity, volume — scalar triple product.
Parallel test — a × b = 0 or a = λb.
Perpendicular test — a · b = 0.

Sanity checks

|a · b| ≤ |a||b| (Cauchy–Schwarz). If your dot product is bigger, you have a computation error.
l² + m² + n² = 1 for direction cosines — always.
|a × b|² + (a · b)² = |a|²|b|². Elegant identity, useful as a mistake radar.