How do I choose dot product or cross product?

Samacheer Kalvi 12th Maths Chapter 6 Exercise Solutions

Chapter 6 — Exercise Solutions Guide

Vector problems are direct if you first decide whether the question needs a scalar answer or a vector answer.

How to Use This Page

1. Start with Key Concepts to refresh the core formulas and definitions. 2. Attempt MCQs, Fill in the Blanks, and True/False to test your recall of basic facts. 3. Try Match the Following to associate each formula with its geometric meaning. 4. Use Answer Briefly (2-mark) for short-answer board exam practice. 5. Study Answer in Detail (5-mark) for full-step solutions examiners expect. 6. Review Common Mistakes before mock tests to avoid sign and formula errors. 7. Check the Exam Checklist the night before your exam.

Key Concepts (Quick Recap)

Term	Definition
Dot product (scalar product)	`a · b = a1b1 + a2b2 + a3b3 =	a		b	cos θ`; result is a scalar
Cross product (vector product)	`a × b` gives a vector perpendicular to both a and b; magnitude = `	a		b	sin θ`
Angle between two vectors	`cos θ = (a · b) / (	a		b	)`
Projection of a on b	`(a · b) /	b	`; scalar quantity representing the component of a along b
Perpendicular vectors	Two vectors are perpendicular if `a · b = 0`
Parallel vectors	Two vectors are parallel if `a × b = 0`
Scalar triple product	`[a b c] = a · (b × c)`; equals the volume of the parallelepiped
Vector triple product	`a × (b × c) = b(a · c) − c(a · b)`
Direction cosines	cos α, cos β, cos γ where `cos²α + cos²β + cos²γ = 1`
Area of parallelogram	`	a × b	` where a and b are adjacent side vectors
Area of triangle	`(1/2)	a × b	`
Coplanar vectors	Three vectors are coplanar if their scalar triple product `[a b c] = 0`
Unit vector	`â = a /	a	`

Formula Recap

Need	Formula
Dot product	a · b = a1b1 + a2b2 + a3b3
Angle	cos theta = (a · b) / (	a		b	)
Cross product magnitude		a x b	=	a		b	sin theta
Area of parallelogram		a x b
Area of triangle	1/2	a x b
Projection of a on b	(a · b) /	b

Standard Solution Pattern

1. Write the given vectors in component form. 2. Identify dot product or cross product. 3. Substitute components carefully. 4. Use square root only at magnitude step. 5. Write units if area or length is asked.

Worked Pattern

For vectors:

a = i + 2j + 3k
b = 2i - j + k

Dot product:

a · b = (1)(2) + (2)(-1) + (3)(1) = 3

Area Pattern

If triangle has adjacent side vectors a and b:

Area = 1/2 |a x b|

Do not forget the 1/2.

I. Choose the Best Answer (MCQs)

1. If a · b = 0, then the angle between a and b is:

(Dot product zero means vectors are perpendicular, so angle = 90°.)

(a) 0°
(b) 45°
(c) 90°
(d) 180°

2. The magnitude of a × b equals:

(a) |a||b| cos θ
(b) |a||b| sin θ
(c) a · b
(d) |a| + |b|

*(Cross product magnitude =	a		b	sin θ.)*

3. If a = (1, 2, 3) and b = (4, 5, 6), then a · b equals:

(a · b = 1(4) + 2(5) + 3(6) = 4 + 10 + 18 = 32.)

(a) 15
(b) 20
(c) 32
(d) 36

4. The area of the parallelogram with sides a = (1, 0, 0) and b = (0, 1, 0) is:

(a) 0
(b) 1
(c) 2
(d) √2

*(a × b = (0, 0, 1);	a × b	= 1.)*

5. The scalar triple product [a b c] equals:

(The scalar triple product is invariant under cyclic permutation.)

(a) a · (b × c)
(b) b · (c × a)
(c) c · (a × b)
(d) All of the above

6. If [a b c] = 0, then the vectors are:

(Zero scalar triple product means the vectors lie in the same plane.)

(a) Mutually perpendicular
(b) Coplanar
(c) Parallel
(d) Unit vectors

7. Projection of a = (3, 4, 0) on b = (1, 0, 0) is:

(a) 4
(b) 3
(c) 5
(d) 7

*(Projection = (a · b)/	b	= 3/1 = 3.)*

8. i × j equals:

(i × j = k by the right-hand rule.)

(a) i
(b) j
(c) k
(d) -k

9. a × a equals:

(Cross product of a vector with itself is always the zero vector.)

(a) |a|²
(b) a²
(c) 0
(d) |a|

10. If direction cosines of a vector are l, m, n, then l² + m² + n² equals:

(Sum of squares of direction cosines is always 1.)

(a) 0
(b) 2
(c) 1
(d) 3

11. The vector triple product a × (b × c) equals:

(This is the BAC − CAB rule: a × (b × c) = b(a · c) − c(a · b).)

(a) b(a · c) + c(a · b)
(b) b(a · c) − c(a · b)
(c) c(a · b) − b(a · c)
(d) a(b · c)

II. Fill in the Blanks

1. The dot product of two perpendicular vectors is ______.

Answer0

2. If `a = (2, 3, 1)`, then `	a	` = ______. Answer: √14

3. The cross product a × b is ______ to both a and b.

Answerperpendicular

4. The volume of a parallelepiped with edges a, b, c is ______. **Answer:	[a b c]	=	a · (b × c)	**

5. i × i = ______.

Answer0 (zero vector)

6. If a = (1, 1, 1) and b = (2, 2, 2), then a × b = ______.

Answer(0, 0, 0) (parallel vectors)

7. The angle between i and k is ______.

Answer90° (π/2)

8. The area of triangle with vertices at (0,0,0), (1,0,0), and (0,1,0) is ______.

Answer1/2

9. j × k = ______.

Answeri

10. If cos²α + cos²β + cos²γ = 1, then α, β, γ are called ______.

Answerdirection angles

III. True or False

1. Dot product of two vectors is always a scalar. True.

2. Cross product of two vectors is always a scalar. False. Correction: Cross product of two vectors is always a vector, not a scalar.

3. If a × b = 0, then the vectors are parallel. True. (Or one of them is the zero vector.)

4. The scalar triple product changes sign when two vectors are swapped. True. (Swapping two vectors reverses the sign of the determinant.)

5. Projection of a vector on itself equals its magnitude. True. (Since `a · a /	a	=	a	² /	a	=	a	`.)

6. a × b = b × a for all vectors. False. Correction: a × b = −(b × a); the cross product is anti-commutative.

7. Three vectors are coplanar if their scalar triple product is zero. True.

8. The direction cosines of a vector can all be negative simultaneously. False. Correction: While individual direction cosines can be negative, the identity l² + m² + n² = 1 still holds. However, direction cosines are the cosines of angles with positive axes, and they can be negative for obtuse angles, but they cannot all be greater than zero in magnitude than allowed by the constraint.

IV. Match the Following

Column A	Column B
1. a · b = 0	(d) Vectors are perpendicular
2. a × b = 0	(e) Vectors are parallel
3. [a b c] = 0	(c) Vectors are coplanar
4. Projection of a on b	(g) (a · b)/	b
5.	a × b		(a) Area of parallelogram
6. (1/2)	a × b		(f) Area of triangle
7. a × (b × c)	(b) b(a · c) − c(a · b)
8. l² + m² + n²	(h) = 1 (direction cosines property)

V. Answer Briefly (2-mark)

1. Find the angle between the vectors a = (1, 1, 0) and b = (0, 1, 1).

a · b = (1)(0) + (1)(1) + (0)(1) = 1
|a| = √(1 + 1 + 0) = √2
|b| = √(0 + 1 + 1) = √2
cos θ = (a · b)/(|a||b|) = 1/(√2 · √2) = 1/2
θ = 60° (π/3)

The angle between the vectors is 60°.

2. Find a × b where a = (1, 2, 3) and b = (4, 5, 6).

a × b = | i   j   k |
        | 1   2   3 |
        | 4   5   6 |
= i(12 − 15) − j(6 − 12) + k(5 − 8)
= (−3, 6, −3)

The cross product is (−3, 6, −3).

3. Show that a = (2, −1, 3) and b = (4, −2, 6) are parallel.

a × b = | i    j   k |
        | 2   −1   3 |
        | 4   −2   6 |
= i(−6 − (−6)) − j(12 − 12) + k(−4 − (−4))
= i(0) − j(0) + k(0) = (0, 0, 0)

Since a × b = 0, the vectors are parallel.

4. Find the projection of a = (4, 3) on b = (2, 0).

a · b = (4)(2) + (3)(0) = 8
|b| = √(4 + 0) = 2
Projection = (a · b)/|b| = 8/2 = 4

The projection of a on b is 4.

5. Find the direction cosines of the vector (1, 2, 2).

|a| = √(1 + 4 + 4) = √9 = 3
l = 1/3, m = 2/3, n = 2/3
Check: l² + m² + n² = 1/9 + 4/9 + 4/9 = 9/9 = 1 ✓

The direction cosines are (1/3, 2/3, 2/3).

VI. Answer in Detail (5-mark)

1. Find the area of the triangle formed by the points A(1, 2, 3), B(2, 3, 1), and C(3, 1, 2).

Step 1: Find two side vectors.

AB = B − A = (2−1, 3−2, 1−3) = (1, 1, −2)
AC = C − A = (3−1, 1−2, 2−3) = (2, −1, −1)

Step 2: Find the cross product AB × AC.

AB × AC = | i    j    k |
           | 1    1   −2 |
           | 2   −1   −1 |
= i(1(−1) − (−2)(−1)) − j(1(−1) − (−2)(2)) + k(1(−1) − 1(2))
= i(−1 − 2) − j(−1 + 4) + k(−1 − 2)
= (−3, −3, −3)

Step 3: Find the magnitude.

|AB × AC| = √(9 + 9 + 9) = √27 = 3√3

Step 4: Area of triangle.

Area = (1/2)|AB × AC| = (1/2)(3√3) = 3√3/2

The area of the triangle is 3√3/2 square units.

2. Prove that the points A(1, 0, −1), B(2, 1, 0), and C(0, 1, −2) are collinear using vector methods.

If the points are collinear, the vectors AB and AC must be parallel (i.e., one is a scalar multiple of the other).

AB = (2−1, 1−0, 0−(−1)) = (1, 1, 1)
AC = (0−1, 1−0, −2−(−1)) = (−1, 1, −1)

Check if AB × AC = 0:

AB × AC = | i    j    k |
           | 1    1    1 |
           | −1   1   −1 |
= i(1(−1) − 1(1)) − j(1(−1) − 1(−1)) + k(1(1) − 1(−1))
= i(−2) − j(0) + k(2)
= (−2, 0, 2) ≠ 0

Since AB × AC ≠ 0, the vectors are not parallel. So the three points are not collinear.

Wait — let us re-check by another method. Let us verify with the scalar triple product approach or check if AB = k × BC.

BC = (0−2, 1−1, −2−0) = (−2, 0, −2)

Is AB a scalar multiple of BC? AB = (1, 1, 1) and BC = (−2, 0, −2). No, since 1/(−2) ≠ 1/0. The points are indeed not collinear.

Hence, by the cross product method, the points A, B, C are not collinear. The area of the triangle formed is `(1/2)	AB × AC	= (1/2)√(4+0+4) = (1/2)√8 = √2`.

3. Find the volume of the parallelepiped with coterminous edges a = (1, 0, 0), b = (1, 1, 0), and c = (1, 1, 1).

Volume =	scalar triple product [a b c]	=	a · (b × c)	.

Step 1: Find b × c.

b × c = | i   j   k |
        | 1   1   0 |
        | 1   1   1 |
= i(1 − 0) − j(1 − 0) + k(1 − 1)
= (1, −1, 0)

Step 2: Find a · (b × c).

a · (b × c) = (1)(1) + (0)(−1) + (0)(0) = 1

Volume =	1	= 1 cubic unit.

4. If a = (2, −1, 1) and b = (1, 1, 2), find a unit vector perpendicular to both a and b.

Step 1: Find a × b.

a × b = | i    j   k |
        | 2   −1   1 |
        | 1    1   2 |
= i((−1)(2) − (1)(1)) − j((2)(2) − (1)(1)) + k((2)(1) − (−1)(1))
= i(−2 − 1) − j(4 − 1) + k(2 + 1)
= (−3, −3, 3)

Step 2: Find the magnitude.

|a × b| = √(9 + 9 + 9) = √27 = 3√3

Step 3: Unit vector.

n̂ = (a × b)/|a × b| = (−3, −3, 3)/(3√3) = (−1, −1, 1)/√3
= (−1/√3, −1/√3, 1/√3)

The unit vector perpendicular to both a and b is (−1/√3) i + (−1/√3) j + (1/√3) k.

5. Find the angle between the vectors a = (3, 4, 5) and b = (1, 0, 0).

a · b = (3)(1) + (4)(0) + (5)(0) = 3
|a| = √(9 + 16 + 25) = √50 = 5√2
|b| = √1 = 1
cos θ = (a · b)/(|a||b|) = 3/(5√2 × 1) = 3/(5√2) = 3√2/10
θ = cos⁻¹(3√2/10)

The angle between the vectors is cos⁻¹(3√2/10).

Worked Examples

Example 1 — Dot product and perpendicularity

Show that a = (2, 3, 1) and b = (1, −1, 1) are perpendicular.

a · b = (2)(1) + (3)(−1) + (1)(1) = 2 − 3 + 1 = 0
Dot product = 0 → the vectors are perpendicular.

Example 2 — Cross product

Find a × b for a = (1, 2, 3) and b = (4, 5, 6).

a × b = | i   j   k |
        | 1   2   3 |
        | 4   5   6 |
= i(2·6 − 3·5) − j(1·6 − 3·4) + k(1·5 − 2·4)
= i(12 − 15) − j(6 − 12) + k(5 − 8)
= (−3, 6, −3)

Example 3 — Scalar triple product (volume)

Find the volume of the parallelepiped with edges a = (1,0,0), b = (0,2,0), c = (0,0,3).

Volume = |a · (b × c)| = | 1 0 0; 0 2 0; 0 0 3 | = 1(2·3 − 0) = 6 cubic units

Common Mistakes to Avoid

❌ Using dot product when area is asked → ✅ Use cross product for area (|a × b| for parallelogram, 1/2|a × b| for triangle)
❌ Forgetting unit vector symbols in vector answer → ✅ A unit vector must have magnitude 1; always divide by |v| to get v̂
❌ Missing negative signs in determinant expansion for cross product → ✅ Write the full determinant and expand carefully; watch the alternating signs
❌ Writing projection without dividing by magnitude → ✅ Projection of a on b = (a · b)/|b|, not just (a · b)
❌ Confusing scalar triple product with vector triple product → ✅ [a b c] = a · (b × c) is a scalar; a × (b × c) is a vector
❌ Using |a × b| = |a||b| cos θ → ✅ |a × b| = |a||b| sin θ; the cos formula is for dot product
❌ Forgetting the 1/2 factor in triangle area → ✅ Area of triangle = (1/2)|a × b|, not |a × b|
❌ Assuming cross product is commutative → ✅ a × b = −(b × a); it is anti-commutative

Exam Checklist

Product type is correct (dot for scalar, cross for vector)
Components are substituted in order
Cross product uses determinant method with correct signs
Magnitude is non-negative (always take √ of sum of squares)
For angle: use cos θ = (a · b)/(|a||b|), not sin θ
For area: use cross product, not dot product
For projection: divide by |b|, not by |a|
Scalar triple product: write as determinant for easy evaluation
Coplanarity check: [a b c] = 0
Final answer matches what the question asks (scalar or vector)
Direction cosines: verify l² + m² + n² = 1

Frequently Asked Questions

Use dot product for angle, projection, perpendicular checks, and scalar value. Use cross product for area, vector perpendicular to two vectors, and parallelogram or triangle area.