Question 1

Let **a** = 2**i** + **j** + **k**, **b** = **i** − 2**j** + 3**k**, and **c** = 3**i** + **j** − 2**k** be three vectors. If **p** = **a** + λ**b** is perpendicular to **c**, then the value of λ is:

Accepted Answer

We need to find λ such that **p** ⊥ **c**, which means **p** · **c** = 0.

Step 1: Express **p** in terms of λ.
**p** = **a** + λ**b**
**p** = (2**i** + **j** + **k**) + λ(**i** − 2**j** + 3**k**)
**p** = (2 + λ)**i** + (1 − 2λ)**j** + (1 + 3λ)**k**

Step 2: Use the perpendicularity condition **p** · **c** = 0.
**c** = 3**i** + **j** − 2**k**

**p** · **c** = (2 + λ)(3) + (1 − 2λ)(1) + (1 + 3λ)(−2) = 0

Step 3: Expand and simplify.
3(2 + λ) + (1 − 2λ) − 2(1 + 3λ) = 0
6 + 3λ + 1 − 2λ − 2 − 6λ = 0
5 − 5λ = 0
5λ = 5
λ = 1

Step 4: Verification.
When λ = 1, **p** = 3**i** − **j** + 4**k**
**p** · **c** = 3(3) + (−1)(1) + 4(−2) = 9 − 1 − 8 = 0 ✓

The condition of perpendicularity (dot product equals zero) is the key theorem applied here.

Question 2

Let **a** = 2**i** + 3**j** + **k** and **b** = **i** − 2**j** + 3**k**. The magnitude of the vector **a** × **b** is:

Accepted Answer

To find |**a** × **b**|, we compute the cross product.

**a** × **b** = |**i**  **j**  **k**|
              |2   3   1|
              |1  −2   3|

= **i**(3·3 − 1·(−2)) − **j**(2·3 − 1·1) + **k**(2·(−2) − 3·1)
= **i**(9 + 2) − **j**(6 − 1) + **k**(−4 − 3)
= 11**i** − 5**j** − 7**k**

Magnitude: |**a** × **b**| = √(11² + (−5)² + (−7)²) = √(121 + 25 + 49) = √195

Alternatively, using |**a** × **b**| = |**a**||**b**|sin θ:
|**a**| = √(4 + 9 + 1) = √14
|**b**| = √(1 + 4 + 9) = √14
**a** · **b** = 2(1) + 3(−2) + 1(3) = 2 − 6 + 3 = −1
cos θ = −1/14, so sin² θ = 1 − 1/196 = 195/196
|**a** × **b**| = √14 · √14 · √(195/196) = 14 · √195/14 = √195

Question 3

Two vectors **u** and **v** satisfy |**u**| = 5, |**v**| = 3, and **u** · **v** = 6. The angle θ between **u** and **v** is:

Accepted Answer

The dot product formula relates the angle between two vectors:
**u** · **v** = |**u**||**v**|cos θ

Substituting the given values:
6 = 5 · 3 · cos θ
6 = 15 cos θ
cos θ = 6/15 = 2/5

Therefore, θ = arccos(2/5).

Numerically: 2/5 = 0.4, so θ ≈ 66.42°, which is approximately 66° 25' or in radians ≈ 1.159 rad.

The exact answer is θ = arccos(2/5).

Question 4

If **a** = **i** + 2**j** + 3**k**, **b** = 2**i** − **j** + **k**, and **c** = **i** + **j** − **k**, then the scalar triple product **a** · (**b** × **c**) equals:

Accepted Answer

The scalar triple product **a** · (**b** × **c**) can be computed as the determinant:

**a** · (**b** × **c**) = |1   2   3|
                        |2  −1   1|
                        |1   1  −1|

Expanding along the first row:
= 1 · |−1   1| − 2 · |2   1| + 3 · |2  −1|
      | 1  −1|       |1  −1|       |1   1|

= 1 · [(−1)(−1) − (1)(1)] − 2 · [(2)(−1) − (1)(1)] + 3 · [(2)(1) − (−1)(1)]
= 1 · [1 − 1] − 2 · [−2 − 1] + 3 · [2 + 1]
= 1 · 0 − 2 · (−3) + 3 · 3
= 0 + 6 + 9
= 15

Question 5

The direction cosines of the vector **v** = 3**i** + 4**j** are:

Accepted Answer

Direction cosines are the cosines of the angles that a vector makes with the positive x, y, and z axes.

For a vector **v** = a**i** + b**j** + c**k**, the direction cosines are:
l = a/|**v**|, m = b/|**v**|, n = c/|**v**|

For **v** = 3**i** + 4**j** + 0**k**:
|**v**| = √(3² + 4² + 0²) = √(9 + 16) = √25 = 5

Direction cosines:
l = 3/5
m = 4/5
n = 0/5 = 0

Therefore, the direction cosines are (3/5, 4/5, 0).

Note: Direction cosines always satisfy l² + m² + n² = 1.
Verification: (3/5)² + (4/5)² + 0² = 9/25 + 16/25 = 25/25 = 1 ✓

Question 6

Let **p** = 2**i** − **j** + **k** and **q** = **i** + 3**j** − 2**k**. The component of **p** along **q** is:

Accepted Answer

The component (or projection) of vector **p** along vector **q** is given by:
Comp_**q** **p** = (**p** · **q**) / |**q**|

Step 1: Compute **p** · **q**
**p** · **q** = (2)(1) + (−1)(3) + (1)(−2)
           = 2 − 3 − 2
           = −3

Step 2: Compute |**q**|
|**q**| = √(1² + 3² + (−2)²)
       = √(1 + 9 + 4)
       = √14

Step 3: Compute the component
Comp_**q** **p** = −3 / √14 = −3√14 / 14

Alternatively, this can be expressed as −3/√14 or −3√14/14 (rationalized form).

Question 7

Let **a** = 2**i** + 3**j** + **k** and **b** = **i** − 2**j** + 3**k**. The magnitude of the vector **a** − **b** is:

Accepted Answer

Setup: We need to find the magnitude of the difference of two vectors using the component subtraction rule and the magnitude formula.

Solution:
Step 1: Compute **a** − **b**
**a** − **b** = (2**i** + 3**j** + **k**) − (**i** − 2**j** + 3**k**)
= (2−1)**i** + (3−(−2))**j** + (1−3)**k**
= **i** + 5**j** − 2**k**

Step 2: Apply the magnitude formula |**v**| = √(v₁² + v₂² + v₃²)
|**a** − **b**| = √(1² + 5² + (−2)²)
= √(1 + 25 + 4)
= √30

The correct answer is √30.

Question 8

If **u** = 3**i** − **j** + 2**k** and **v** = **i** + 2**j** − **k**, then the dot product **u** · **v** equals:

Accepted Answer

Setup: Dot product of two vectors is computed using the formula **u** · **v** = u₁v₁ + u₂v₂ + u₃v₃.

Solution:
Step 1: Identify components
**u** = (3, −1, 2)
**v** = (1, 2, −1)

Step 2: Apply the dot product formula
**u** · **v** = (3)(1) + (−1)(2) + (2)(−1)
= 3 − 2 − 2
= −1

The correct answer is −1.

Question 9

Two vectors **p** and **q** are such that |**p**| = 4, |**q**| = 3, and **p** · **q** = 6. The angle θ between **p** and **q** is:

Accepted Answer

Setup: The angle between two vectors is found using the dot product formula: **p** · **q** = |**p**| |**q**| cos θ.

Solution:
Step 1: Apply the dot product angle formula
**p** · **q** = |**p**| |**q**| cos θ
6 = (4)(3) cos θ
6 = 12 cos θ

Step 2: Solve for cos θ
cos θ = 6/12 = 1/2

Step 3: Find θ
θ = arccos(1/2) = 60° or π/3 radians

The correct answer is 60° (or π/3 radians).

Question 10

Let **a** = **i** + **j** + **k** and **b** = 2**i** − **j** + 3**k**. The cross product **a** × **b** is:

Accepted Answer

Setup: The cross product of two vectors is computed using the determinant formula:
**a** × **b** = |**i**  **j**  **k**|
              |a₁  a₂  a₃|
              |b₁  b₂  b₃|

Solution:
Step 1: Set up the determinant
**a** × **b** = |**i**  **j**  **k**|
              |1   1   1|
              |2  −1   3|

Step 2: Expand along the first row
**a** × **b** = **i**(1·3 − 1·(−1)) − **j**(1·3 − 1·2) + **k**(1·(−1) − 1·2)
= **i**(3 + 1) − **j**(3 − 2) + **k**(−1 − 2)
= 4**i** − **j** − 3**k**

The correct answer is 4**i** − **j** − 3**k**.

Question 11

If **m** = 2**i** + **j** − **k** and **n** = **i** − 2**j** + 2**k**, then the magnitude of **m** × **n** is:

Accepted Answer

Setup: The magnitude of a cross product can be computed as |**m** × **n**| = |**m**| |**n**| sin θ, or by computing the cross product explicitly and finding its magnitude.

Solution:
Step 1: Compute **m** × **n** using the determinant
**m** × **n** = |**i**  **j**  **k**|
              |2   1  −1|
              |1  −2   2|

Step 2: Expand along the first row
**m** × **n** = **i**(1·2 − (−1)·(−2)) − **j**(2·2 − (−1)·1) + **k**(2·(−2) − 1·1)
= **i**(2 − 2) − **j**(4 + 1) + **k**(−4 − 1)
= 0**i** − 5**j** − 5**k**
= −5**j** − 5**k**

Step 3: Compute the magnitude
|**m** × **n**| = √(0² + (−5)² + (−5)²)
= √(0 + 25 + 25)
= √50
= 5√2

The correct answer is 5√2.

Question 12

If **u** = 3**i** − **j** + 2**k** and **v** = **i** + 2**j** − **k**, then **u** · **v** equals:

Accepted Answer

Setup: The dot product of two vectors is computed by multiplying corresponding components and summing the results.

Solution:
Step 1: Identify components
**u** = (3, −1, 2)
**v** = (1, 2, −1)

Step 2: Apply the dot product formula **u** · **v** = u₁v₁ + u₂v₂ + u₃v₃
**u** · **v** = (3)(1) + (−1)(2) + (2)(−1)
= 3 − 2 − 2
= −1

The correct answer is −1.

Question 13

The direction cosines of the vector **v** = 4**i** − 4**j** + 2**k** are:

Accepted Answer

Setup: Direction cosines are the cosines of the angles a vector makes with the positive x, y, and z axes. They are found by dividing each component by the magnitude of the vector.

Solution:
Step 1: Compute the magnitude of **v**
|**v**| = √(4² + (−4)² + 2²)
= √(16 + 16 + 4)
= √36
= 6

Step 2: Divide each component by the magnitude
Direction cosines (l, m, n) are:
l = 4/6 = 2/3
m = −4/6 = −2/3
n = 2/6 = 1/3

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

Question 14

The angle θ between vectors **p** = **i** + **j** and **q** = **j** + **k** is given by cos θ =:

Accepted Answer

Setup: The angle between two vectors is found using the dot product formula: cos θ = (**p** · **q**) / (|**p**| |**q**|).

Solution:
Step 1: Compute the dot product **p** · **q**
**p** · **q** = (1)(0) + (1)(1) + (0)(1)
= 0 + 1 + 0
= 1

Step 2: Compute the magnitudes
|**p**| = √(1² + 1² + 0²) = √2
|**q**| = √(0² + 1² + 1²) = √2

Step 3: Apply the angle formula
cos θ = 1 / (√2 · √2)
= 1 / 2

The correct answer is 1/2.

Question 15

If **a** and **b** are two vectors such that |**a**| = 5, |**b**| = 3, and **a** · **b** = 12, then |**a** + **b**|² equals:

Accepted Answer

Setup: The magnitude squared of a vector sum is found using the expansion |**a** + **b**|² = |**a**|² + 2(**a** · **b**) + |**b**|², which follows from the definition of the dot product.

Solution:
Step 1: Identify given values
|**a**| = 5, so |**a**|² = 25
|**b**| = 3, so |**b**|² = 9
**a** · **b** = 12

Step 2: Apply the expansion formula
|**a** + **b**|² = |**a**|² + 2(**a** · **b**) + |**b**|²
= 25 + 2(12) + 9
= 25 + 24 + 9
= 58

The correct answer is 58.

Question 16

The angle between the vectors **a** = 2**i** + **j** − 2**k** and **b** = **i** + **j** + **k** is θ. Then cos θ equals:

Accepted Answer

Setup: The angle θ between two vectors is found using cos θ = (**a** · **b**)/(|**a**||**b**|).

Step 1: Compute **a** · **b**:
**a** · **b** = (2)(1) + (1)(1) + (−2)(1) = 2 + 1 − 2 = 1

Step 2: Compute |**a**|:
|**a**| = √(4 + 1 + 4) = √9 = 3

Step 3: Compute |**b**|:
|**b**| = √(1 + 1 + 1) = √3

Step 4: Compute cos θ:
cos θ = 1/(3√3) = 1/(3√3) × (√3/√3) = √3/9

Step 5: Verify:
cos θ = √3/9 ≈ 0.1925, which corresponds to θ ≈ 78.9°, a reasonable angle.

Question 17

The vectors **a** = **i** + 2**j** + 3**k**, **b** = 2**i** + **j** − **k**, and **c** = 3**i** + 3**j** + 2**k** form a scalar triple product. Calculate **a** · (**b** × **c**).

Accepted Answer

The scalar triple product **a** · (**b** × **c**) can be computed as the determinant:

**a** · (**b** × **c**) = |1   2   3 |
                        |2   1  −1 |
                        |3   3   2 |

Expanding along the first row:
= 1(1·2 − (−1)·3) − 2(2·2 − (−1)·3) + 3(2·3 − 1·3)
= 1(2 + 3) − 2(4 + 3) + 3(6 − 3)
= 1(5) − 2(7) + 3(3)
= 5 − 14 + 9
= 0

The scalar triple product is 0, which means the three vectors are coplanar.

Question 18

A line passes through the point P(1, 2, 3) and is parallel to the vector **d** = 2**i** + **j** − 2**k**. Find the distance from the point Q(2, 3, 4) to this line.

Accepted Answer

The distance from a point Q to a line passing through point P with direction vector **d** is given by:

distance = |(**PQ**) × **d**| / |**d**|

Step 1: Find **PQ**:
**PQ** = Q − P = (2−1)**i** + (3−2)**j** + (4−3)**k** = **i** + **j** + **k**

Step 2: Compute **PQ** × **d**:
**PQ** × **d** = |**i**  **j**  **k**|
               |1    1    1 |
               |2    1   −2 |

= **i**(1·(−2) − 1·1) − **j**(1·(−2) − 1·2) + **k**(1·1 − 1·2)
= **i**(−2 − 1) − **j**(−2 − 2) + **k**(1 − 2)
= −3**i** + 4**j** − **k**

Step 3: Find |**PQ** × **d**|:
|**PQ** × **d**| = √((−3)² + 4² + (−1)²) = √(9 + 16 + 1) = √26

Step 4: Find |**d**|:
|**d**| = √(2² + 1² + (−2)²) = √(4 + 1 + 4) = √9 = 3

Step 5: Calculate distance:
distance = √26 / 3

Question 19

Let **p** and **q** be two vectors such that |**p**| = 5, |**q**| = 12, and |**p** + **q**| = 13. Find the angle θ between **p** and **q**.

Accepted Answer

We use the formula for the magnitude of a vector sum:
|**p** + **q**|² = |**p**|² + 2(**p** · **q**) + |**q**|²

Substituting the given values:
13² = 5² + 2(**p** · **q**) + 12²
169 = 25 + 2(**p** · **q**) + 144
169 = 169 + 2(**p** · **q**)
0 = 2(**p** · **q**)
**p** · **q** = 0

Since **p** · **q** = |**p**| |**q**| cos θ:
0 = 5 · 12 · cos θ
0 = 60 cos θ
cos θ = 0

Therefore: θ = 90° or π/2 radians

Note: This is a Pythagorean triple (5, 12, 13), which confirms that **p** and **q** are perpendicular.

Question 20

Let **a** = 2**i** + 3**j** + **k** and **b** = **i** − 2**j** + 3**k**. Find the magnitude of the vector **a** × **b**.

Accepted Answer

To find the magnitude of the cross product, we first compute **a** × **b** using the determinant formula.

**a** × **b** = |**i**  **j**  **k**|
              |2    3    1|
              |1   −2    3|

= **i**(3·3 − 1·(−2)) − **j**(2·3 − 1·1) + **k**(2·(−2) − 3·1)
= **i**(9 + 2) − **j**(6 − 1) + **k**(−4 − 3)
= 11**i** − 5**j** − 7**k**

Magnitude: |**a** × **b**| = √(11² + (−5)² + (−7)²) = √(121 + 25 + 49) = √195

Simplifying: √195 = √(195) = √(3·5·13) — cannot simplify further.

However, let me verify: 121 + 25 + 49 = 195. Since 195 = 3 × 65 = 3 × 5 × 13, the answer is √195.

Question 21

The vectors **a** = **i** + 2**j** + 3**k**, **b** = 2**i** + **j** − **k**, and **c** = 7**i** + 5**j** + **k** are coplanar if and only if their scalar triple product equals zero. Which of the following statements is correct?

Accepted Answer

Three vectors are coplanar if and only if their scalar triple product **a** · (**b** × **c**) = 0, which is equivalent to the determinant:

|1   2   3|
|2   1  −1| = 0
|7   5   1|

Expanding along the first row:
= 1(1·1 − (−1)·5) − 2(2·1 − (−1)·7) + 3(2·5 − 1·7)
= 1(1 + 5) − 2(2 + 7) + 3(10 − 7)
= 1(6) − 2(9) + 3(3)
= 6 − 18 + 9
= −3

Since the scalar triple product is −3 ≠ 0, the vectors are NOT coplanar.

Therefore, the correct statement is that the vectors are NOT coplanar.

CDS Vectors

CDS Vectors — Past Exam Questions

Vectors— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Vectors — Practice Questions

60-Second Revision — Vectors