Vectors — Class 12 Maths practice questions

Question 1

A drone is being tested on a flat airfield. Its displacement from the launch pad is recorded at two checkpoints:

At Checkpoint P, the position vector is p⃗ = 3î + 4ĵ + 5k̂ (in metres).
At Checkpoint Q, the position vector is q⃗ = 7î + 2ĵ + 8k̂ (in metres).

A ground sensor is placed at point S with position vector s⃗ = 5î + 3ĵ + 6.5k̂.

(i) Find the displacement vector PQ⃗ of the drone from P to Q.   [1]
(ii) Find the unit vector in the direction of PQ⃗.   [1]
(iii) The drone's signal strength is proportional to the scalar projection of PS⃗ onto PQ⃗. Find this scalar projection.   [2]
        OR
(iii) Verify whether S lies on the line segment PQ. If yes, find the ratio PS : SQ.   [2]

Accepted Answer

(i) PQ⃗ = q⃗ − p⃗

PQ⃗ = (7î + 2ĵ + 8k̂) − (3î + 4ĵ + 5k̂)

∴ PQ⃗ = 4î − 2ĵ + 3k̂

(ii) |PQ⃗| = √(4² + (−2)² + 3²) = √(16 + 4 + 9) = √29

Unit vector P̂Q = PQ⃗ / |PQ⃗|

∴ P̂Q = (1/√29)(4î − 2ĵ + 3k̂)

(iii) PS⃗ = s⃗ − p⃗

PS⃗ = (5î + 3ĵ + 6.5k̂) − (3î + 4ĵ + 5k̂) = 2î − ĵ + 1.5k̂

Scalar projection of PS⃗ onto PQ⃗ = (PS⃗ · PQ⃗) / |PQ⃗|

PS⃗ · PQ⃗ = (2)(4) + (−1)(−2) + (1.5)(3) = 8 + 2 + 4.5 = 14.5

∴ Scalar projection = 14.5 / √29 = 29 / (2√29) = √29 / 2

∴ The scalar projection of PS⃗ onto PQ⃗ is √29 / 2 metres.

OR

(iii) If S lies on segment PQ, then PS⃗ = λ · PQ⃗ for some λ ∈ (0, 1).

PS⃗ = 2î − ĵ + 1.5k̂  and  PQ⃗ = 4î − 2ĵ + 3k̂

Comparing î components: 2 = 4λ ⟹ λ = 1/2

Checking ĵ component: −1 = −2(1/2) = −1  ✓

Checking k̂ component: 1.5 = 3(1/2) = 1.5  ✓

∵ λ = 1/2 ∈ (0, 1) and all three components are consistent, S lies on segment PQ.

PS : SQ = λ : (1 − λ) = (1/2) : (1/2)

∴ S lies on PQ and divides it in the ratio PS : SQ = 1 : 1 (S is the midpoint of PQ).

Question 2

A drone delivery company uses GPS coordinates to plan flight paths. Three relay towers are located at positions A(1, 2, 3), B(3, 0, 1) and C(2, 4, 5) in a 3D coordinate grid (units in km). A drone flies from tower A to tower B, and a separate signal beam is directed from tower A toward tower C.

(i) Find the vector AB⃗ and the vector AC⃗.
(ii) Find AB⃗ × AC⃗ (the cross product).
(iii) Hence find the area of the triangular region covered by the three relay towers. Also determine whether the three towers are collinear or non-collinear, justifying your answer.

Accepted Answer

(i) Finding AB⃗ and AC⃗:  [1 mark]

AB⃗ = OB⃗ − OA⃗ = (3î + 0ĵ + k̂) − (î + 2ĵ + 3k̂)

∴ AB⃗ = 2î − 2ĵ − 2k̂

AC⃗ = OC⃗ − OA⃗ = (2î + 4ĵ + 5k̂) − (î + 2ĵ + 3k̂)

∴ AC⃗ = î + 2ĵ + 2k̂

(ii) Finding AB⃗ × AC⃗:  [1 mark]

AB⃗ × AC⃗ = |î  ĵ  k̂|
              |2  −2  −2|
              |1   2   2|

= î[(−2)(2) − (−2)(2)] − ĵ[(2)(2) − (−2)(1)] + k̂[(2)(2) − (−2)(1)]

= î[−4 + 4] − ĵ[4 + 2] + k̂[4 + 2]

= î(0) − ĵ(6) + k̂(6)

∴ AB⃗ × AC⃗ = 0î − 6ĵ + 6k̂

(iii) Area of triangular region and collinearity:  [2 marks]

|AB⃗ × AC⃗| = √(0² + (−6)² + 6²) = √(0 + 36 + 36) = √72 = 6√2

Area of △ABC = ½|AB⃗ × AC⃗| = ½ × 6√2

∴ Area of the triangular region covered by the three relay towers = 3√2 sq. km

Since AB⃗ × AC⃗ = 0î − 6ĵ + 6k̂ ≠ 0⃗,

∴ The three relay towers A, B and C are non-collinear.

(Note: If the cross product were 0⃗, the vectors would be parallel, implying collinearity. Since it is non-zero, the three points form a proper triangle.)

Question 3

A civil engineer is designing a triangular park in a city layout. The park has vertices at points A, B and C whose position vectors are

OA⃗ = î + 2ĵ + 3k̂,   OB⃗ = 3î − ĵ + k̂,   OC⃗ = 2î + 3ĵ − k̂

Based on the above information, answer the following questions:

(i) Find the vector AB⃗.
(ii) Find the vector AC⃗.
(iii) Find the area of the triangular park. Express your answer in sq. units.

OR

(iii) A landscaper wants to place a decorative rope along two sides AB and AC of the park. Verify whether AB⃗ and AC⃗ are perpendicular to each other. Hence, if they are perpendicular, find the area of the triangular park directly using the formula for the area of a right triangle.

Accepted Answer

(i) AB⃗ = OB⃗ − OA⃗

∴ AB⃗ = (3î − ĵ + k̂) − (î + 2ĵ + 3k̂)

∴ AB⃗ = 2î − 3ĵ − 2k̂         [1 mark]

(ii) AC⃗ = OC⃗ − OA⃗

∴ AC⃗ = (2î + 3ĵ − k̂) − (î + 2ĵ + 3k̂)

∴ AC⃗ = î + ĵ − 4k̂         [1 mark]

(iii) Area of △ABC = ½|AB⃗ × AC⃗|

Computing AB⃗ × AC⃗ using the determinant formula:

AB⃗ × AC⃗ = |  î    ĵ    k̂  |
             |  2   −3   −2  |
             |  1    1   −4  |

= î[(−3)(−4) − (−2)(1)] − ĵ[(2)(−4) − (−2)(1)] + k̂[(2)(1) − (−3)(1)]

= î[12 + 2] − ĵ[−8 + 2] + k̂[2 + 3]

= 14î + 6ĵ + 5k̂

|AB⃗ × AC⃗| = √(14² + 6² + 5²)
             = √(196 + 36 + 25)
             = √257

∴ Area of the triangular park = ½√257 sq. units         [2 marks]

━━━━━━━━━━━━━━━━━━━━
OR
━━━━━━━━━━━━━━━━━━━━

(iii) To check whether AB⃗ ⊥ AC⃗, we compute the dot product.

AB⃗ · AC⃗ = (2)(1) + (−3)(1) + (−2)(−4)
           = 2 − 3 + 8
           = 7

∵ AB⃗ · AC⃗ = 7 ≠ 0, AB⃗ is NOT perpendicular to AC⃗.

∴ The two sides AB and AC of the park are not perpendicular to each other, so the right-triangle area formula ½ × base × height (with AB and AC as legs) is NOT directly applicable here.

However, we may still compute the area using the cross-product method.

|AB⃗| = √(2² + (−3)² + (−2)²) = √(4 + 9 + 4) = √17

|AC⃗| = √(1² + 1² + (−4)²) = √(1 + 1 + 16) = √18 = 3√2

From part (iii) main: |AB⃗ × AC⃗| = √257

∴ Area of the triangular park = ½|AB⃗ × AC⃗| = ½√257 sq. units

∴ Since AB⃗ and AC⃗ are not perpendicular, the direct right-triangle formula does not apply; the area of the triangular park is ½√257 sq. units.         [2 marks]

Question 4

A civil engineer is designing a triangular park in a city layout. The park has vertices at points A, B and C. Using a coordinate grid, the position vectors of A, B and C are taken as:

OA⃗ = 2î + 3ĵ + k̂,   OB⃗ = 4î + ĵ + 2k̂,   OC⃗ = 6î + 5ĵ + 3k̂

Based on the above information, answer the following questions:

(i) Find the vector AB⃗.

(ii) Find the vector AC⃗.

(iii) Find the area of the triangular park. Hence determine whether the three vertices A, B, C are collinear.

OR

(iii) Find a unit vector perpendicular to both AB⃗ and AC⃗.

Accepted Answer

(i) AB⃗ = OB⃗ − OA⃗

⟹ AB⃗ = (4î + ĵ + 2k̂) − (2î + 3ĵ + k̂)

∴ AB⃗ = 2î − 2ĵ + k̂

[1 mark]

(ii) AC⃗ = OC⃗ − OA⃗

⟹ AC⃗ = (6î + 5ĵ + 3k̂) − (2î + 3ĵ + k̂)

∴ AC⃗ = 4î + 2ĵ + 2k̂

[1 mark]

(iii) Area of △ABC = ½|AB⃗ × AC⃗|

Computing AB⃗ × AC⃗:

AB⃗ × AC⃗ = |î   ĵ   k̂  |
             |2  −2   1  |
             |4   2   2  |

= î[(−2)(2) − (1)(2)] − ĵ[(2)(2) − (1)(4)] + k̂[(2)(2) − (−2)(4)]

= î[−4 − 2] − ĵ[4 − 4] + k̂[4 + 8]

= −6î − 0ĵ + 12k̂

⟹ AB⃗ × AC⃗ = −6î + 12k̂

|AB⃗ × AC⃗| = √[(−6)² + 0² + 12²] = √[36 + 0 + 144] = √180 = 6√5

∴ Area of triangular park = ½ × 6√5 = 3√5 sq. units

∵ Area = 3√5 ≠ 0, the three vertices A, B, C are NOT collinear.

[2 marks]

OR

(iii) A unit vector perpendicular to both AB⃗ and AC⃗ is along AB⃗ × AC⃗.

From the working above, AB⃗ × AC⃗ = −6î + 0ĵ + 12k̂

|AB⃗ × AC⃗| = √[(−6)² + 0² + 12²] = √[36 + 144] = √180 = 6√5

∴ Required unit vector = (AB⃗ × AC⃗) / |AB⃗ × AC⃗|

= (−6î + 12k̂) / 6√5

∴ Required unit vector = (−î + 2k̂) / √5

[2 marks]

Question 5

A structural engineer is designing a triangular support frame for a bridge. The three vertices of the triangular frame are located at points A(1, 2, 3), B(3, 0, 1), and C(2, 4, 0) in space (coordinates in metres).

(i) Find the vectors AB⃗ and AC⃗.
(ii) Find the area of the triangular frame, correct to two decimal places.
(iii) A force F⃗ = 3î − ĵ + 2k̂ (in Newtons) acts at vertex A. Find the scalar projection of F⃗ along the direction of AB⃗.
    OR
    Find a unit vector perpendicular to the plane of the triangular frame.

Accepted Answer

(i) AB⃗ = OB⃗ − OA⃗ = (3î + 0ĵ + k̂) − (î + 2ĵ + 3k̂) = 2î − 2ĵ − 2k̂

AC⃗ = OC⃗ − OA⃗ = (2î + 4ĵ + 0k̂) − (î + 2ĵ + 3k̂) = î + 2ĵ − 3k̂

∴ AB⃗ = 2î − 2ĵ − 2k̂ and AC⃗ = î + 2ĵ − 3k̂           [1 mark]

(ii) Area of △ABC = ½|AB⃗ × AC⃗|

AB⃗ × AC⃗ = |î  ĵ  k̂|
             |2  −2  −2|
             |1   2  −3|

= î[(−2)(−3) − (−2)(2)] − ĵ[(2)(−3) − (−2)(1)] + k̂[(2)(2) − (−2)(1)]

= î[6 + 4] − ĵ[−6 + 2] + k̂[4 + 2]

= 10î + 4ĵ + 6k̂

|AB⃗ × AC⃗| = √(10² + 4² + 6²) = √(100 + 16 + 36) = √152 = 2√38

∴ Area of triangular frame = ½ × 2√38 = √38 ≈ 6.16 sq. metres           [1 mark]

(iii) Scalar projection of F⃗ along AB⃗ = (F⃗ · AB⃗) / |AB⃗|

F⃗ · AB⃗ = (3î − ĵ + 2k̂) · (2î − 2ĵ − 2k̂)
        = (3)(2) + (−1)(−2) + (2)(−2)
        = 6 + 2 − 4 = 4

|AB⃗| = √(2² + (−2)² + (−2)²) = √(4 + 4 + 4) = √12 = 2√3

∴ Scalar projection of F⃗ along AB⃗ = 4 / (2√3) = 2/√3 = (2√3)/3 N           [2 marks]

OR

A unit vector perpendicular to the plane of the triangular frame is along AB⃗ × AC⃗.

From part (ii): AB⃗ × AC⃗ = 10î + 4ĵ + 6k̂ and |AB⃗ × AC⃗| = 2√38

∴ Required unit vector n̂ = (AB⃗ × AC⃗) / |AB⃗ × AC⃗|

n̂ = (10î + 4ĵ + 6k̂) / (2√38)

n̂ = (5î + 2ĵ + 3k̂) / √38

∴ Required unit vector perpendicular to the plane of the frame = (5î + 2ĵ + 3k̂)/√38           [2 marks]

Question 6

A drone is flying in a warehouse. Its position at time t = 0 is given by point A with position vector a⃗ = 2î + 3ĵ + k̂, and it moves to point B with position vector b⃗ = 4î − ĵ + 5k̂. A sensor is placed at point M, the midpoint of AB. A second drone starts from point C with position vector c⃗ = −2î + 5ĵ − 3k̂.

Based on the above information, answer the following:

(i) Find the position vector of the midpoint M of AB.

(ii) Find the vector AB⃗.

(iii) Find the vector CM⃗. Hence determine the unit vector along CM⃗.

OR

(iii) Find the area of triangle ABC.

Accepted Answer

(i) The position vector of the midpoint M of segment AB is given by:

OM⃗ = (a⃗ + b⃗)/2

∴ OM⃗ = [(2î + 3ĵ + k̂) + (4î − ĵ + 5k̂)] / 2

= (6î + 2ĵ + 6k̂) / 2

∴ Position vector of M = 3î + ĵ + 3k̂    [1 mark]

(ii) AB⃗ = b⃗ − a⃗

= (4î − ĵ + 5k̂) − (2î + 3ĵ + k̂)

∴ AB⃗ = 2î − 4ĵ + 4k̂    [1 mark]

(iii) CM⃗ = OM⃗ − OC⃗

= (3î + ĵ + 3k̂) − (−2î + 5ĵ − 3k̂)

= (3 + 2)î + (1 − 5)ĵ + (3 + 3)k̂

∴ CM⃗ = 5î − 4ĵ + 6k̂

|CM⃗| = √(5² + (−4)² + 6²) = √(25 + 16 + 36) = √77

∴ Unit vector along CM⃗ = CM⃗/|CM⃗| = (5î − 4ĵ + 6k̂)/√77    [2 marks]

OR

(iii) AB⃗ = 2î − 4ĵ + 4k̂  [from part (ii)]

AC⃗ = c⃗ − a⃗ = (−2î + 5ĵ − 3k̂) − (2î + 3ĵ + k̂) = −4î + 2ĵ − 4k̂

AB⃗ × AC⃗ = |î  ĵ  k̂|
              |2  −4  4|
              |−4  2  −4|

= î[(−4)(−4) − (4)(2)] − ĵ[(2)(−4) − (4)(−4)] + k̂[(2)(2) − (−4)(−4)]

= î[16 − 8] − ĵ[−8 + 16] + k̂[4 − 16]

= 8î − 8ĵ − 12k̂

|AB⃗ × AC⃗| = √(8² + (−8)² + (−12)²) = √(64 + 64 + 144) = √272 = 4√17

∴ Area of △ABC = (1/2)|AB⃗ × AC⃗| = (1/2) × 4√17 = 2√17 sq. units    [2 marks]

Question 7

A drone delivery company models its flight paths using vectors. Three relay towers are located at positions A(1, 2, 3), B(3, 4, 5) and C(2, 0, 1) in a coordinate grid (units in km).

(i) Find the vector AB⃗ and AC⃗.

(ii) Find the unit vector along AB⃗.

(iii) A delivery drone must fly from A along a path perpendicular to both AB⃗ and AC⃗. Find the vector along which the drone must travel.

OR

(iii) Find the area of triangle ABC formed by the three relay towers.

Accepted Answer

(i) AB⃗ = OB⃗ − OA⃗ = (3î + 4ĵ + 5k̂) − (î + 2ĵ + 3k̂)

∴ AB⃗ = 2î + 2ĵ + 2k̂

AC⃗ = OC⃗ − OA⃗ = (2î + 0ĵ + k̂) − (î + 2ĵ + 3k̂)

∴ AC⃗ = î − 2ĵ − 2k̂

(ii) |AB⃗| = √(2² + 2² + 2²) = √(4 + 4 + 4) = √12 = 2√3

Unit vector along AB⃗ = AB⃗ / |AB⃗| = (2î + 2ĵ + 2k̂) / (2√3)

∴ Unit vector along AB⃗ = (1/√3)î + (1/√3)ĵ + (1/√3)k̂

(iii) The drone must travel along a direction perpendicular to both AB⃗ and AC⃗, i.e., along AB⃗ × AC⃗.

AB⃗ × AC⃗ = |î  ĵ  k̂ |
              |2  2  2 |
              |1  −2  −2|

= î[(2)(−2) − (2)(−2)] − ĵ[(2)(−2) − (2)(1)] + k̂[(2)(−2) − (2)(1)]

= î[−4 + 4] − ĵ[−4 − 2] + k̂[−4 − 2]

= î(0) − ĵ(−6) + k̂(−6)

∴ AB⃗ × AC⃗ = 6ĵ − 6k̂

∴ The drone must travel along the vector 6ĵ − 6k̂ (or equivalently along ĵ − k̂).

OR

(iii) Area of triangle ABC = ½|AB⃗ × AC⃗|

From above, AB⃗ × AC⃗ = 6ĵ − 6k̂

|AB⃗ × AC⃗| = √(0² + 6² + (−6)²) = √(0 + 36 + 36) = √72 = 6√2

Area of △ABC = ½ × 6√2

∴ Area of △ABC = 3√2 sq. km

Question 8

A drone delivery company models the flight path of two drones using vectors. Drone A travels along the vector a⃗ = 4î − 3ĵ + 2k̂ (in km), and Drone B travels along the vector b⃗ = 2î + 4ĵ − k̂ (in km), both measured from the same base station.

Based on this information, answer the following:

(i) Find the unit vector in the direction of the total displacement a⃗ + b⃗.

(ii) Find the angle between the flight paths of Drone A and Drone B. (Express as cos⁻¹ of an exact value.)

(iii) Find the area of the triangle formed by the origin O and the tips of vectors a⃗ and b⃗ (i.e., the triangle OAB), correct to the form k√m sq. km.
   OR
   Find the scalar projection of Drone A's displacement on Drone B's displacement, and hence find the component of a⃗ perpendicular to b⃗.

Accepted Answer

(i) [1 mark]

Total displacement: a⃗ + b⃗ = (4+2)î + (−3+4)ĵ + (2−1)k̂ = 6î + ĵ + k̂

|a⃗ + b⃗| = √(6² + 1² + 1²) = √(36 + 1 + 1) = √38

∴ Required unit vector = (6î + ĵ + k̂)/√38

(ii) [1 mark]

|a⃗| = √(4² + (−3)² + 2²) = √(16 + 9 + 4) = √29

|b⃗| = √(2² + 4² + (−1)²) = √(4 + 16 + 1) = √21

a⃗ · b⃗ = (4)(2) + (−3)(4) + (2)(−1) = 8 − 12 − 2 = −6

Using cosθ = (a⃗ · b⃗)/(|a⃗||b⃗|),

cosθ = −6/(√29 · √21) = −6/√609

∴ Angle between the flight paths = cos⁻¹(−6/√609)

(iii) [2 marks]

Area of triangle OAB = ½|a⃗ × b⃗|

Computing a⃗ × b⃗ using the determinant:

a⃗ × b⃗ = |î  ĵ  k̂ |
             |4  −3  2|
             |2   4 −1|

= î[(−3)(−1) − (2)(4)] − ĵ[(4)(−1) − (2)(2)] + k̂[(4)(4) − (−3)(2)]

= î[3 − 8] − ĵ[−4 − 4] + k̂[16 + 6]

= −5î + 8ĵ + 22k̂

|a⃗ × b⃗| = √((−5)² + 8² + 22²) = √(25 + 64 + 484) = √573

∴ Area of triangle OAB = ½√573 sq. km

OR

Scalar projection of a⃗ on b⃗ = (a⃗ · b⃗)/|b⃗| = −6/√21

∴ Scalar projection of Drone A's displacement on Drone B's displacement = −6/√21 km

Vector projection of a⃗ on b⃗ = ((a⃗ · b⃗)/|b⃗|²)b⃗ = (−6/21)(2î + 4ĵ − k̂) = (−2/7)(2î + 4ĵ − k̂)

= −4î/7 − 8ĵ/7 + 2k̂/7

Component of a⃗ perpendicular to b⃗ = a⃗ − (vector projection of a⃗ on b⃗)

= (4î − 3ĵ + 2k̂) − (−4î/7 − 8ĵ/7 + 2k̂/7)

= (4 + 4/7)î + (−3 + 8/7)ĵ + (2 − 2/7)k̂

= (32/7)î + (−13/7)ĵ + (12/7)k̂

∴ Component of a⃗ perpendicular to b⃗ = (1/7)(32î − 13ĵ + 12k̂)

Question 9

If |a⃗| = 3, |b⃗| = 4 and a⃗ · b⃗ = 6, then the angle between a⃗ and b⃗ is :

Accepted Answer

Option (C) is correct.

Explanation: Using a⃗ · b⃗ = |a⃗||b⃗|cosθ, we get 6 = 3 × 4 × cosθ ⟹ cosθ = 6/12 = 1/2 ⟹ θ = π/3.

Question 10

If |a⃗| = 5 and |b⃗| = 4 and a⃗ · b⃗ = 0, then |a⃗ × b⃗| is equal to:

Accepted Answer

Option (C) is correct.

Explanation: Since a⃗ · b⃗ = 0, the vectors a⃗ and b⃗ are perpendicular, so the angle θ between them is π/2.

∴ |a⃗ × b⃗| = |a⃗| |b⃗| sin θ = 5 × 4 × sin(π/2) = 5 × 4 × 1 = 20.

Question 11

If a⃗ = 3î − ĵ + 2k̂ and b⃗ = î + 2ĵ − k̂, find the value of λ such that (a⃗ + λb⃗) is perpendicular to b⃗.

Accepted Answer

We have a⃗ = 3î − ĵ + 2k̂ and b⃗ = î + 2ĵ − k̂.

a⃗ + λb⃗ = (3 + λ)î + (−1 + 2λ)ĵ + (2 − λ)k̂

For (a⃗ + λb⃗) ⊥ b⃗, we need (a⃗ + λb⃗) · b⃗ = 0.

⟹ (3 + λ)(1) + (−1 + 2λ)(2) + (2 − λ)(−1) = 0

⟹ 3 + λ − 2 + 4λ − 2 + λ = 0

⟹ 6λ − 1 = 0

∴ λ = 1/6

Question 12

If â is a unit vector and (x⃗ − â)·(x⃗ + â) = 8, then find |x⃗|.

Accepted Answer

Given: â is a unit vector, so |â| = 1, and (x⃗ − â)·(x⃗ + â) = 8.

Expanding the dot product,

|x⃗|² − |â|² = 8

⟹ |x⃗|² − 1 = 8

⟹ |x⃗|² = 9

∴ |x⃗| = 3

Question 13

A vector r⃗ makes equal angles with all three coordinate axes. If the magnitude of r⃗ is 4√3 units, find r⃗.

Accepted Answer

Since r⃗ makes equal angles with all three coordinate axes, its direction cosines satisfy

l = m = n.

∵ cos²α + cos²β + cos²γ = 1

⟹ l² + l² + l² = 1  ⟹  3l² = 1  ⟹  l = ±1/√3

∴ The unit vector along r⃗ is r̂ = ±(1/√3)(î + ĵ + k̂).

Given |r⃗| = 4√3, we have

r⃗ = |r⃗| · r̂ = 4√3 · (±1/√3)(î + ĵ + k̂)

∴ r⃗ = ±4(î + ĵ + k̂)

Question 14

If the vectors a⃗ = 3î − ĵ + 2k̂ and b⃗ = î + 2ĵ − k̂ are perpendicular to each other, find the value of λ, given that a⃗ = λî − ĵ + 2k̂ and b⃗ = î + 2ĵ − k̂, where a⃗ · b⃗ = 0.

Accepted Answer

Given: a⃗ = λî − ĵ + 2k̂ and b⃗ = î + 2ĵ − k̂ are perpendicular to each other.

For perpendicular vectors, a⃗ · b⃗ = 0.

a⃗ · b⃗ = (λ)(1) + (−1)(2) + (2)(−1)

⟹ λ − 2 − 2 = 0

⟹ λ − 4 = 0

∴ λ = 4

Question 15

Find a vector of magnitude 7 units which is perpendicular to both the vectors a⃗ = 2î − 3ĵ + 6k̂ and b⃗ = î + ĵ − k̂.

Accepted Answer

We need a vector perpendicular to both a⃗ and b⃗, so we first compute a⃗ × b⃗.

a⃗ × b⃗ = |î  ĵ  k̂|
            |2  −3   6|
            |1   1  −1|

= î[(−3)(−1) − (6)(1)] − ĵ[(2)(−1) − (6)(1)] + k̂[(2)(1) − (−3)(1)]

= î[3 − 6] − ĵ[−2 − 6] + k̂[2 + 3]

= −3î + 8ĵ + 5k̂

∴ a⃗ × b⃗ = −3î + 8ĵ + 5k̂  ...(i)

Magnitude of a⃗ × b⃗:
|a⃗ × b⃗| = √((−3)² + (8)² + (5)²)
          = √(9 + 64 + 25)
          = √98
          = 7√2  ...(ii)

The required unit vector perpendicular to both a⃗ and b⃗ is:
(a⃗ × b⃗) / |a⃗ × b⃗| = (−3î + 8ĵ + 5k̂) / (7√2)

The required vector of magnitude 7 units is:
7 × (a⃗ × b⃗) / |a⃗ × b⃗| = 7 × (−3î + 8ĵ + 5k̂) / (7√2)

= (−3î + 8ĵ + 5k̂) / √2

= (−3/√2)î + (8/√2)ĵ + (5/√2)k̂

∴ Required vector = (−3î + 8ĵ + 5k̂) / √2  (or equivalently (−3√2/2)î + 4√2 ĵ + (5√2/2)k̂)

Question 16

The two adjacent sides of a parallelogram are represented by the vectors a⃗ = 3î − ĵ + 2k̂ and b⃗ = î + 3ĵ − k̂. Find the unit vector parallel to its diagonal (a⃗ + b⃗). Also, find the area of the parallelogram.

Accepted Answer

Given: a⃗ = 3î − ĵ + 2k̂ and b⃗ = î + 3ĵ − k̂

Step 1: Find the diagonal vector d⃗ = a⃗ + b⃗.

d⃗ = (3+1)î + (−1+3)ĵ + (2−1)k̂ = 4î + 2ĵ + k̂

Step 2: Find |d⃗|.

|d⃗| = √(4² + 2² + 1²) = √(16 + 4 + 1) = √21

Step 3: Write the unit vector parallel to d⃗.

d̂ = d⃗/|d⃗| = (1/√21)(4î + 2ĵ + k̂)

∴ Unit vector parallel to diagonal = (4î + 2ĵ + k̂)/√21

Step 4: Compute a⃗ × b⃗ to find the area.

a⃗ × b⃗ = |î  ĵ  k̂ |
             |3  −1  2|
             |1   3  −1|

= î[(−1)(−1) − (2)(3)] − ĵ[(3)(−1) − (2)(1)] + k̂[(3)(3) − (−1)(1)]

= î[1 − 6] − ĵ[−3 − 2] + k̂[9 + 1]

= −5î + 5ĵ + 10k̂

Step 5: Find |a⃗ × b⃗|.

|a⃗ × b⃗| = √((−5)² + 5² + 10²) = √(25 + 25 + 100) = √150 = 5√6

∴ Area of parallelogram = |a⃗ × b⃗| = 5√6 sq. units

Question 17

Find a unit vector perpendicular to each of the vectors (a⃗ + b⃗) and (a⃗ − b⃗), where a⃗ = 3î + 2ĵ + 2k̂ and b⃗ = î + 2ĵ − 2k̂.

Accepted Answer

Given: a⃗ = 3î + 2ĵ + 2k̂ and b⃗ = î + 2ĵ − 2k̂.

Step 1: Find (a⃗ + b⃗) and (a⃗ − b⃗).

a⃗ + b⃗ = (3+1)î + (2+2)ĵ + (2−2)k̂ = 4î + 4ĵ + 0k̂

a⃗ − b⃗ = (3−1)î + (2−2)ĵ + (2+2)k̂ = 2î + 0ĵ + 4k̂

Step 2: Find (a⃗ + b⃗) × (a⃗ − b⃗).

A vector perpendicular to both (a⃗ + b⃗) and (a⃗ − b⃗) is their cross product.

Let p⃗ = (a⃗ + b⃗) × (a⃗ − b⃗)

p⃗ = |  î    ĵ    k̂  |
      |  4    4    0  |
      |  2    0    4  |

p⃗ = î[(4)(4) − (0)(0)] − ĵ[(4)(4) − (0)(2)] + k̂[(4)(0) − (4)(2)]

p⃗ = î[16 − 0] − ĵ[16 − 0] + k̂[0 − 8]

p⃗ = 16î − 16ĵ − 8k̂

Step 3: Find |p⃗|.

|p⃗| = √(16² + (−16)² + (−8)²)
     = √(256 + 256 + 64)
     = √576
     = 24

Step 4: Find the required unit vector.

Required unit vector p̂ = p⃗/|p⃗|

p̂ = (16î − 16ĵ − 8k̂) / 24

∴ Required unit vector = (2/3)î − (2/3)ĵ − (1/3)k̂

Question 18

The two adjacent sides of a parallelogram are represented by the vectors a⃗ = 2î − 4ĵ + 5k̂ and b⃗ = î − 2ĵ − 3k̂. Find the vectors representing the diagonals of the parallelogram and hence find the length of the longer diagonal.

Accepted Answer

Let the parallelogram ABCD have AB⃗ = a⃗ = 2î − 4ĵ + 5k̂ and AD⃗ = b⃗ = î − 2ĵ − 3k̂.

The diagonals of the parallelogram are AC⃗ and BD⃗, where:

AC⃗ = a⃗ + b⃗

AC⃗ = (2î − 4ĵ + 5k̂) + (î − 2ĵ − 3k̂)

AC⃗ = 3î − 6ĵ + 2k̂

BD⃗ = b⃗ − a⃗

BD⃗ = (î − 2ĵ − 3k̂) − (2î − 4ĵ + 5k̂)

BD⃗ = −î + 2ĵ − 8k̂

Length of diagonal AC⃗:

|AC⃗| = √(3² + (−6)² + 2²)

|AC⃗| = √(9 + 36 + 4)

|AC⃗| = √49 = 7

Length of diagonal BD⃗:

|BD⃗| = √((−1)² + 2² + (−8)²)

|BD⃗| = √(1 + 4 + 64)

|BD⃗| = √69

∵ √69 > 7, the longer diagonal is BD⃗.

∴ The diagonals of the parallelogram are AC⃗ = 3î − 6ĵ + 2k̂ and BD⃗ = −î + 2ĵ − 8k̂, and the length of the longer diagonal is √69 units.

Question 19

In a triangle ABC, the position vectors of vertices A, B and C are 2î − ĵ + k̂, 3î + 2ĵ − k̂ and î − 3ĵ + 2k̂ respectively. Find the vector joining A to the mid-point M of BC. Hence find the area of triangle ABM.

Accepted Answer

Let O be the origin.

Given: OA⃗ = 2î − ĵ + k̂,  OB⃗ = 3î + 2ĵ − k̂,  OC⃗ = î − 3ĵ + 2k̂.

Step 1: Position vector of mid-point M of BC.

OM⃗ = (OB⃗ + OC⃗)/2 = [(3î + 2ĵ − k̂) + (î − 3ĵ + 2k̂)] / 2

= (4î − ĵ + k̂) / 2

= 2î − (1/2)ĵ + (1/2)k̂

Step 2: Vector AM⃗.

AM⃗ = OM⃗ − OA⃗

= [2î − (1/2)ĵ + (1/2)k̂] − [2î − ĵ + k̂]

= (2−2)î + (−1/2+1)ĵ + (1/2−1)k̂

= (1/2)ĵ − (1/2)k̂

Step 3: Vector AB⃗.

AB⃗ = OB⃗ − OA⃗

= (3î + 2ĵ − k̂) − (2î − ĵ + k̂)

= î + 3ĵ − 2k̂

Step 4: Area of triangle ABM = (1/2)|AB⃗ × AM⃗|.

AB⃗ × AM⃗ = |  î    ĵ    k̂  |
             |  1    3   −2  |
             |  0   1/2 −1/2 |

= î[(3)(−1/2) − (−2)(1/2)] − ĵ[(1)(−1/2) − (−2)(0)] + k̂[(1)(1/2) − (3)(0)]

= î[−3/2 + 1] − ĵ[−1/2 − 0] + k̂[1/2 − 0]

= î(−1/2) − ĵ(−1/2) + k̂(1/2)

= −(1/2)î + (1/2)ĵ + (1/2)k̂

|AB⃗ × AM⃗| = √[(−1/2)² + (1/2)² + (1/2)²]

= √[1/4 + 1/4 + 1/4]

= √(3/4)

= (√3)/2

∴ Area of triangle ABM = (1/2) × (√3)/2 = (√3)/4 sq. units

Question 20

Let $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$. Find a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. Also find the sine of the angle between $\vec{a}$ and $\vec{b}$.

Accepted Answer

Given: $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.

**Step 1: Compute $\vec{a} 	imes \vec{b}$**

$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 3 \ 1 & 2 & -1 \end{vmatrix}$

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

$= \hat{i}[1 - 6] - \hat{j}[-2 - 3] + \hat{k}[4 + 1]$

$= -5\hat{i} + 5\hat{j} + 5\hat{k}$ ...(i)

**Step 2: Find $|\vec{a} 	imes \vec{b}|$**

$|\vec{a} 	imes \vec{b}| = \sqrt{(-5)^2 + (5)^2 + (5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}$ ...(ii)

**Step 3: Required unit vector perpendicular to both $\vec{a}$ and $\vec{b}$**

A unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is $\hat{n} = \dfrac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|}$.

From (i) and (ii),

$\hat{n} = \dfrac{-5\hat{i} + 5\hat{j} + 5\hat{k}}{5\sqrt{3}} = \dfrac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$

$	herefore$ Required unit vector $= \pm \dfrac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$

**Step 4: Find $\sin	heta$ using $|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}|\sin	heta$**

$|\vec{a}| = \sqrt{(2)^2 + (-1)^2 + (3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$

$|\vec{b}| = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

$	herefore \sin	heta = \dfrac{|\vec{a} 	imes \vec{b}|}{|\vec{a}||\vec{b}|} = \dfrac{5\sqrt{3}}{\sqrt{14} \cdot \sqrt{6}} = \dfrac{5\sqrt{3}}{\sqrt{84}} = \dfrac{5\sqrt{3}}{2\sqrt{21}}$

$= \dfrac{5\sqrt{3}}{2\sqrt{21}} = \dfrac{5\sqrt{3}}{2\sqrt{21}} 	imes \dfrac{\sqrt{21}}{\sqrt{21}} = \dfrac{5\sqrt{63}}{2 	imes 21} = \dfrac{5 	imes 3\sqrt{7}}{42} = \dfrac{15\sqrt{7}}{42} = \dfrac{5\sqrt{7}}{14}$

$	herefore \sin	heta = \dfrac{5\sqrt{7}}{14}$

Vectors: Class 12 Maths Practice Questions

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