Class 12 Maths Chapter 10 Important Questions – Vector Algebra

Vector Algebra studies quantities that have both magnitude and direction, along with operations such as vector addition, dot product and cross product. The chapter covers vector components, direction cosines, section formula, projection, angles, collinearity, perpendicularity and areas.

Important Questions for Class 12 Maths Chapter 10 Vector Algebra help students practise the main concepts covered in the current NCERT textbook. These include different types of vectors, unit vectors, direction cosines, vector addition, section formula, scalar product and vector product.

Students can use these Class 12 Vector Algebra Important Questions with Solutions to revise formulas, understand each calculation and practise questions of different difficulty levels.

Key Takeaways

• A vector has both magnitude and direction.
• A unit vector has a magnitude of one unit.
• The dot product is used to calculate angles, projections and perpendicularity.
• The cross product is used to calculate perpendicular vectors and areas.
• Two vectors are collinear when one is a scalar multiple of the other.

Important Vector Algebra Formulas

For a vector A:

A = a₁i + a₂j + a₃k

Magnitude of A:

|A| = √(a₁² + a₂² + a₃²)

Unit vector in the direction of A:

 = A / |A|

For points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):

PQ = (x₂ − x₁)i + (y₂ − y₁)j + (z₂ − z₁)k

If R divides the line joining P and Q internally in the ratio m:n:

R = (mQ + nP) / (m + n)

Dot product:

A · B = |A||B| cos θ

Angle between two vectors:

cos θ = (A · B) / (|A||B|)

Scalar projection of A on B:

Projection of A on B = (A · B) / |B|

Cross product:

|A × B| = |A||B| sin θ

Area of a triangle:

Area = 1/2 |A × B|

Area of a parallelogram:

Area = |A × B|

Direction cosines:

l² + m² + n² = 1

Access Class 12 Maths Chapter 10 Important Questions in 30 Minutes

• First 10 minutes: Types of vectors, magnitude, components and unit vectors
• Next 10 minutes: Direction cosines, section formula, dot product and projection
• Final 10 minutes: Cross product, collinearity, perpendicularity and area-based questions

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Very Short Answer Questions – 1 Mark

Q1. What is a vector?

Answer: A vector is a quantity that has both magnitude and direction.

Q2. What is a scalar quantity?

Answer: A scalar is a quantity that has magnitude but no direction.

Distance, mass, speed, time and temperature are examples of scalar quantities.

Q3. What is a zero vector?

Answer: A vector whose initial and terminal points coincide is called a zero vector. Its magnitude is zero.

Q4. What is a unit vector?

Answer: A vector whose magnitude is one unit is called a unit vector.

Q5. What are equal vectors?

Answer: Two vectors are equal when they have the same magnitude and direction, irrespective of the positions of their initial points.

Q6. What are collinear vectors?

Answer: Two or more vectors are collinear if they are parallel to the same line.

Q7. What are coinitial vectors?

Answer: Two or more vectors having the same initial point are called coinitial vectors.

Q8. Find the magnitude of A = 2i − 3j + k.

Solution:

|A| = √[2² + (−3)² + 1²]

|A| = √(4 + 9 + 1)

|A| = √14

Q9. Find the unit vector in the direction of A = 2i + 3j + k.

Solution:

|A| = √(2² + 3² + 1²)

|A| = √14

 = A / |A|

 = (2i + 3j + k) / √14

Q10. State the relation among direction cosines l, m and n.

Answer:

l² + m² + n² = 1

Q11. When are two non-zero vectors perpendicular?

Answer: Two non-zero vectors A and B are perpendicular when:

A · B = 0

Q12. When are two vectors parallel?

Answer: Two vectors are parallel when one is a scalar multiple of the other.

A = λB

They also satisfy:

A × B = 0

Q13. State the dot products of mutually perpendicular unit vectors.

Answer:

i · j = 0

j · k = 0

k · i = 0

Also:

i · i = j · j = k · k = 1

Q14. State the basic cross-product relations.

Answer:

i × j = k

j × k = i

k × i = j

On reversing the order:

j × i = −k

k × j = −i

i × k = −j

Q15. What is the vector joining points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂)?

Answer:

PQ = (x₂ − x₁)i + (y₂ − y₁)j + (z₂ − z₁)k

Short Answer Questions – 2 or 3 Marks

Q16. Represent graphically a displacement of 40 km, 30° east of north.

Solution:

Draw a north-south line from point O.

From the north direction, measure an angle of 30° towards the east. Draw a directed line segment OP in this direction.

Using the scale:

1 cm = 10 km

The length of OP should be 4 cm.

Therefore, OP represents a displacement of 40 km, 30° east of north.

Q17. Find the vector joining P(2, 3, 0) and Q(−1, −2, −4), directed from P to Q.

Solution:

PQ = (x₂ − x₁)i + (y₂ − y₁)j + (z₂ − z₁)k

PQ = (−1 − 2)i + (−2 − 3)j + (−4 − 0)k

PQ = −3i − 5j − 4k

Q18. Find the magnitude of the vector joining P(2, 3, 0) and Q(−1, −2, −4).

Solution:

PQ = −3i − 5j − 4k

|PQ| = √[(−3)² + (−5)² + (−4)²]

|PQ| = √(9 + 25 + 16)

|PQ| = √50

|PQ| = 5√2

Q19. Find the position vector of the midpoint of the line joining P(2, 3, 4) and Q(4, 1, −2).

Solution:

Midpoint coordinates:

[(2 + 4)/2, (3 + 1)/2, (4 − 2)/2]

= (3, 2, 1)

Therefore, the position vector is:

3i + 2j + k

Q20. Find the position vector of a point R that divides the line joining P and Q internally in the ratio m:n.

Solution:

Let the position vectors of P and Q be P and Q.

Using the internal section formula:

R = (mQ + nP) / (m + n)

Q21. Find the position vector of a point R that divides the line joining P and Q externally in the ratio m:n.

Solution:

Using the external section formula:

R = (mQ − nP) / (m − n)

Here, m ≠ n.

Q22. Find the direction cosines of A = i + j − 2k.

Solution:

The direction ratios are:

1, 1, −2

Magnitude:

|A| = √[1² + 1² + (−2)²]

|A| = √6

Therefore:

l = 1/√6

m = 1/√6

n = −2/√6

Q23. Show that A = i + j + k is equally inclined to the coordinate axes.

Solution:

|A| = √(1² + 1² + 1²)

|A| = √3

Direction cosines:

cos α = 1/√3

cos β = 1/√3

cos γ = 1/√3

Since all three direction cosines are equal, A is equally inclined to the x, y and z axes.

Q24. Find the angle between A = i + j − k and B = −i + j + k.

Solution:

A · B = (1)(−1) + (1)(1) + (−1)(1)

A · B = −1 + 1 − 1

A · B = −1

|A| = √3

|B| = √3

cos θ = (A · B) / (|A||B|)

cos θ = −1 / (√3 × √3)

cos θ = −1/3

Therefore:

θ = cos⁻¹(−1/3)

Q25. Find the projection of A = 2i + 3j + 2k on B = i + 2j + k.

Solution:

A · B = (2)(1) + (3)(2) + (2)(1)

A · B = 2 + 6 + 2

A · B = 10

|B| = √(1² + 2² + 1²)

|B| = √6

Projection of A on B:

Projection = (A · B) / |B|

Projection = 10/√6

Q26. Find a vector of magnitude 7 units in the direction of A = i − 2j.

Solution:

|A| = √[1² + (−2)²]

|A| = √5

Unit vector along A:

 = (i − 2j) / √5

Required vector:

7Â = 7(i − 2j) / √5

Therefore:

Required vector = (7/√5)i − (14/√5)j

Q27. Show that A = 2i − 3j + 4k and B = −4i + 6j − 8k are collinear.

Solution:

B = −2(2i − 3j + 4k)

Therefore:

B = −2A

Since one vector is a scalar multiple of the other, A and B are collinear.

Long Answer Questions – 4 or 5 Marks

Q28. Show that the points with position vectors 2i − j + k, i − 3j − 5k and 3i − 4j − 4k form a right-angled triangle.

Solution:

Let the points be A, B and C.

Position vector of A:

A = 2i − j + k

Position vector of B:

B = i − 3j − 5k

Position vector of C:

C = 3i − 4j − 4k

AB = B − A

AB = (1 − 2)i + (−3 + 1)j + (−5 − 1)k

AB = −i − 2j − 6k

BC = C − B

BC = (3 − 1)i + (−4 + 3)j + (−4 + 5)k

BC = 2i − j + k

CA = A − C

CA = (2 − 3)i + (−1 + 4)j + (1 + 4)k

CA = −i + 3j + 5k

Now:

|AB|² = (−1)² + (−2)² + (−6)²

|AB|² = 1 + 4 + 36

|AB|² = 41

|BC|² = 2² + (−1)² + 1²

|BC|² = 4 + 1 + 1

|BC|² = 6

|CA|² = (−1)² + 3² + 5²

|CA|² = 1 + 9 + 25

|CA|² = 35

Since:

|BC|² + |CA|² = 6 + 35

|BC|² + |CA|² = 41

Therefore:

|AB|² = |BC|² + |CA|²

Hence, triangle ABC is right-angled at C.

Q29. Prove that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, −1) are collinear.

Solution:

AB = (2 − 1)i + (6 − 2)j + (3 − 7)k

AB = i + 4j − 4k

BC = (3 − 2)i + (10 − 6)j + (−1 − 3)k

BC = i + 4j − 4k

Therefore:

AB = BC

Since AB and BC are parallel, the points A, B and C are collinear.

Q30. Find a unit vector perpendicular to both A = i + j − k and B = 2i − j + k.

Solution:

A vector perpendicular to both A and B is A × B.

Using the determinant:

A × B =

| i j k |

| 1 1 −1 |

| 2 −1 1 |

Expanding:

A × B = i[(1)(1) − (−1)(−1)] − j[(1)(1) − (−1)(2)] + k[(1)(−1) − (1)(2)]

A × B = i(1 − 1) − j(1 + 2) + k(−1 − 2)

A × B = −3j − 3k

Magnitude:

|A × B| = √[0² + (−3)² + (−3)²]

|A × B| = √18

|A × B| = 3√2

Required unit vector:

(A × B) / |A × B|

= (−3j − 3k) / (3√2)

= −(j + k) / √2

The vector (j + k)/√2 is also perpendicular to both A and B.

Q31. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Solution:

AB = (2 − 1)i + (3 − 1)j + (5 − 2)k

AB = i + 2j + 3k

AC = (1 − 1)i + (5 − 1)j + (5 − 2)k

AC = 4j + 3k

Now:

AB × AC =

| i j k |

| 1 2 3 |

| 0 4 3 |

Expanding:

AB × AC = i[(2)(3) − (3)(4)] − j[(1)(3) − (3)(0)] + k[(1)(4) − (2)(0)]

AB × AC = −6i − 3j + 4k

Magnitude:

|AB × AC| = √[(−6)² + (−3)² + 4²]

|AB × AC| = √(36 + 9 + 16)

|AB × AC| = √61

Area of triangle ABC:

Area = 1/2 |AB × AC|

Area = √61/2 square units

Q32. Find the area of the parallelogram whose adjacent sides are A = 3i + j + 4k and B = i − j + k.

Solution:

A × B =

| i j k |

| 3 1 4 |

| 1 −1 1 |

Expanding:

A × B = i[(1)(1) − (4)(−1)] − j[(3)(1) − (4)(1)] + k[(3)(−1) − (1)(1)]

A × B = 5i + j − 4k

Magnitude:

|A × B| = √[5² + 1² + (−4)²]

|A × B| = √(25 + 1 + 16)

|A × B| = √42

Therefore, the area of the parallelogram is:

√42 square units

Q33. Prove the triangle inequality |A + B| ≤ |A| + |B|.

Solution:

Consider:

|A + B|² = (A + B) · (A + B)

Using the distributive property:

|A + B|² = A · A + B · B + 2A · B

Therefore:

|A + B|² = |A|² + |B|² + 2A · B

Using the Cauchy–Schwarz inequality:

A · B ≤ |A||B|

Therefore:

|A + B|² ≤ |A|² + |B|² + 2|A||B|

|A + B|² ≤ (|A| + |B|)²

Taking square roots:

|A + B| ≤ |A| + |B|

Hence proved.

Q34. Prove that (A + B) · (A − B) = |A|² − |B|².

Solution:

Using the distributive property:

(A + B) · (A − B)

= A · A − A · B + B · A − B · B

Since the dot product is commutative:

A · B = B · A

Therefore:

−A · B + B · A = 0

Hence:

(A + B) · (A − B) = A · A − B · B

Therefore:

(A + B) · (A − B) = |A|² − |B|²

Q35. Using vectors, prove that parallelograms on the same base and between the same parallels have equal areas.

Solution:

Suppose two parallelograms have the same base represented by vector A.

Let their adjacent sides be represented by B and C.

Since the parallelograms lie between the same parallels, the perpendicular components of B and C with respect to A are equal.

Area of the first parallelogram:

Area₁ = |A × B|

Area of the second parallelogram:

Area₂ = |A × C|

Both cross products represent the same base multiplied by the same perpendicular height.

Therefore:

|A × B| = |A × C|

Hence, the two parallelograms have equal areas.

Additional Practice Questions

1. Find the unit vector in the direction of the vector joining P(1, 2, 3) and Q(4, 5, 6).
2. Find a vector of magnitude 8 units in the direction of 5i − 2j + k.
3. Find the direction cosines of the vector joining A(1, 2, −3) and B(−1, −2, 1), directed from A to B.
4. Find the position vector of a point dividing the line joining P and Q internally in the ratio 2:1.
5. Find the angle between A = i − 2j + 3k and B = 3i − 2j + k.
6. Find the projection of A = i − j on B = i + j.
7. Find a unit vector perpendicular to A = 2i + 3j − k and B = i − 2j + 4k.
8. Find the area of a triangle whose adjacent sides are A = 2i − j + k and B = i + 3j − 2k.
9. Find the area of a parallelogram whose adjacent sides are A = i + 2j + 3k and B = 2i − j + k.
10. Show that the vectors A = 3i − 5j + k and B = −6i + 10j − 2k are collinear.

Important Questions Class 12 Maths Chapter-Wise