Question 1

Let **a** = 2**i** + 3**j** + **k** and **b** = **i** − **j** + 2**k**. The scalar projection of **a** onto **b** is:

Accepted Answer

The scalar projection of vector **a** onto vector **b** is given by the formula: proj_b(**a**) = (**a** · **b**) / |**b**|.

Step 1: Calculate **a** · **b**
**a** · **b** = (2)(1) + (3)(−1) + (1)(2) = 2 − 3 + 2 = 1

Step 2: Calculate |**b**|
|**b**| = √(1² + (−1)² + 2²) = √(1 + 1 + 4) = √6

Step 3: Apply the formula
Scalar projection = 1/√6 = √6/6

The scalar projection represents the length of the projection of **a** in the direction of **b**, which is always computed as the dot product divided by the magnitude of the reference vector.

Question 2

The direction cosines of the line passing through points P(1, 2, 3) and Q(4, 5, 6) are:

Accepted Answer

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

Step 1: Find the direction vector **PQ**
**PQ** = Q − P = (4−1, 5−2, 6−3) = (3, 3, 3)

Step 2: Calculate the magnitude of **PQ**
|**PQ**| = √(3² + 3² + 3²) = √27 = 3√3

Step 3: Divide each component by the magnitude to get direction cosines
Direction cosines = (3/(3√3), 3/(3√3), 3/(3√3)) = (1/√3, 1/√3, 1/√3)

Step 4: Rationalize (optional but standard form)
= (√3/3, √3/3, √3/3)

Direction cosines are the unit vector in the direction of the line. They satisfy the property that the sum of their squares equals 1: (1/√3)² + (1/√3)² + (1/√3)² = 1/3 + 1/3 + 1/3 = 1 ✓

Question 3

If **a** and **b** are two vectors such that |**a**| = 5, |**b**| = 3, and **a** · **b** = 12, then the angle θ between them is:

Accepted Answer

The angle between two vectors is found using the dot product formula: **a** · **b** = |**a**| |**b**| cos θ

Step 1: Substitute the given values
12 = (5)(3) cos θ
12 = 15 cos θ

Step 2: Solve for cos θ
cos θ = 12/15 = 4/5

Step 3: Find θ
θ = arccos(4/5)

Since cos θ = 4/5 > 0, the angle is acute (between 0° and 90°).
Numerically: θ ≈ 36.87° or approximately 0.6435 radians.

The angle between two vectors is uniquely determined by their dot product and magnitudes. The formula cos θ = (**a** · **b**)/(|**a**||**b**|) is fundamental to vector geometry.

Question 4

The vectors **a** = **i** + 2**j** + 3**k** and **b** = 2**i** + 4**j** + 6**k** are:

Accepted Answer

To determine the relationship between two vectors, we check if one is a scalar multiple of the other.

Step 1: Check if **b** = λ**a** for some scalar λ
**b** = 2**i** + 4**j** + 6**k**
**a** = **i** + 2**j** + 3**k**

Step 2: Divide corresponding components
2/1 = 2, 4/2 = 2, 6/3 = 2

Step 3: Conclusion
Since all ratios are equal to 2, we have **b** = 2**a**.

This means **b** is a scalar multiple of **a**, so the vectors are parallel (collinear). Parallel vectors have the same or opposite direction.

Step 4: Verify using cross product
**a** × **b** = **a** × (2**a**) = 2(**a** × **a**) = 2(**0**) = **0**

When the cross product is zero, vectors are parallel. This is the defining property of parallel vectors.

Question 5

The equation of the plane passing through the point (1, 2, 3) and perpendicular to the vector **n** = 2**i** − **j** + 3**k** is:

Accepted Answer

The equation of a plane perpendicular to a normal vector **n** = (a, b, c) and passing through point (x₀, y₀, z₀) is:
a(x − x₀) + b(y − y₀) + c(z − z₀) = 0

Step 1: Identify the given information
Normal vector **n** = (2, −1, 3)
Point on plane: (x₀, y₀, z₀) = (1, 2, 3)

Step 2: Apply the plane equation formula
2(x − 1) − 1(y − 2) + 3(z − 3) = 0

Step 3: Expand
2x − 2 − y + 2 + 3z − 9 = 0
2x − y + 3z − 9 = 0

Step 4: Verify by substituting the point (1, 2, 3)
2(1) − 2 + 3(3) − 9 = 2 − 2 + 9 − 9 = 0 ✓

The normal vector is perpendicular to every vector lying in the plane. The coefficients of x, y, z in the plane equation are exactly the components of the normal vector.

Question 6

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

Accepted Answer

The scalar projection (or component) of **a** onto **b** is given by: comp_**b**(**a**) = (**a** · **b**) / |**b**|.

Step 1: Calculate **a** · **b** = (2)(1) + (3)(−1) + (1)(2) = 2 − 3 + 2 = 1.

Step 2: Calculate |**b**| = √(1² + (−1)² + 2²) = √(1 + 1 + 4) = √6.

Step 3: Scalar projection = 1/√6 = √6/6.

The magnitude of the vector projection is the absolute value of the scalar projection, which is √6/6 (positive in this case).

Question 7

If **u** = **i** + 2**j** − **k** and **v** = 3**i** − **j** + 2**k**, then the magnitude of **u** × **v** is:

Accepted Answer

The cross product **u** × **v** is calculated using the determinant formula:

**u** × **v** = |**i**  **j**  **k**|
              |1   2  −1|
              |3  −1   2|

Step 1: **i** component = (2)(2) − (−1)(−1) = 4 − 1 = 3.
Step 2: **j** component = −[(1)(2) − (−1)(3)] = −[2 + 3] = −5.
Step 3: **k** component = (1)(−1) − (2)(3) = −1 − 6 = −7.

So **u** × **v** = 3**i** − 5**j** − 7**k**.

Step 4: |**u** × **v**| = √(3² + (−5)² + (−7)²) = √(9 + 25 + 49) = √83.

Question 8

Two vectors **a** and **b** satisfy |**a**| = 5, |**b**| = 3, and **a** · **b** = 12. The angle θ between them is:

Accepted Answer

The dot product formula relates the angle between two vectors:
**a** · **b** = |**a**| |**b**| cos(θ)

Step 1: Substitute the given values:
12 = (5)(3) cos(θ)
12 = 15 cos(θ)

Step 2: Solve for cos(θ):
cos(θ) = 12/15 = 4/5

Step 3: Find θ:
θ = arccos(4/5) ≈ 36.87° or θ = arccos(0.8)

Note: The angle is acute since cos(θ) > 0.

Question 9

The scalar triple product (**a** × **b**) · **c** for **a** = **i** + **j**, **b** = **j** + **k**, and **c** = **k** + **i** is:

Accepted Answer

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

(**a** × **b**) · **c** = |1  1  0|
                         |0  1  1|
                         |1  0  1|

Step 1: Expand along the first row:
= 1 · |1  1| − 1 · |0  1| + 0 · |0  1|
      |0  1|       |1  1|       |1  0|

Step 2: Calculate the 2×2 minors:
= 1(1·1 − 1·0) − 1(0·1 − 1·1) + 0
= 1(1) − 1(−1)
= 1 + 1
= 2

Verification using cross product method:
**a** × **b** = |**i**  **j**  **k**| = **i**(1−0) − **j**(0−0) + **k**(0−1) = **i** − **k**
              |1   1   0|
              |0   1   1|

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

Wait, let me recalculate the cross product:
**a** × **b** = **i**(1·1 − 0·1) − **j**(1·1 − 0·0) + **k**(1·1 − 1·0) = **i** − **j** + **k**

(**a** × **b**) · **c** = (**i** − **j** + **k**) · (**k** + **i**) = 1·1 + (−1)·0 + 1·1 = 1 + 1 = 2

Question 10

Let vectors **a** = (1, 2, 3), **b** = (2, -1, 1), and **c** = (1, 1, 1). The scalar triple product [**a** **b** **c**] = **a** · (**b** × **c**) is equal to:

Accepted Answer

The scalar triple product [**a** **b** **c**] = **a** · (**b** × **c**) can be computed using the determinant formula:

[**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) − 2·(2 − 1) + 3·(2 + 1)
= 1·(−2) − 2·(1) + 3·(3)
= −2 − 2 + 9
= 5

The scalar triple product equals 5.

Question 11

Two vectors **u** and **v** satisfy |**u**| = 3, |**v**| = 4, 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 = 3 × 4 × cos θ
6 = 12 cos θ
cos θ = 6/12 = 1/2

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

The angle between the vectors is 60°.

Question 12

A line passes through the point P(1, 2, 3) and has direction vector **d** = (2, −1, 2). The distance from the point Q(4, 3, 5) to this line is:

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: Compute **PQ** = Q − P = (4−1, 3−2, 5−3) = (3, 1, 2)

Step 2: Compute **PQ** × **d**:
**PQ** × **d** = (3, 1, 2) × (2, −1, 2)
= |**i**  **j**  **k**|
  |3    1    2|
  |2   −1    2|

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

Step 3: Compute |**PQ** × **d**| = √(16 + 4 + 25) = √45 = 3√5

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

Step 5: distance = 3√5 / 3 = √5

The distance is √5.

Question 13

Three events A, B, and C are mutually exclusive and exhaustive with P(A) = 1/2, P(B) = 1/3, and P(C) = 1/6. If event D occurs with probability P(D|A) = 1/4, P(D|B) = 1/2, and P(D|C) = 1/3, then P(D) is:

Accepted Answer

Since A, B, and C are mutually exclusive and exhaustive, we use the law of total probability:

P(D) = P(D|A)·P(A) + P(D|B)·P(B) + P(D|C)·P(C)

Step 1: Verify that A, B, C are exhaustive:
P(A) + P(B) + P(C) = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ✓

Step 2: Apply the law of total probability:
P(D) = (1/4)·(1/2) + (1/2)·(1/3) + (1/3)·(1/6)

Step 3: Compute each term:
= 1/8 + 1/6 + 1/18

Step 4: Find common denominator (LCM of 8, 6, 18 is 72):
= 9/72 + 12/72 + 4/72
= 25/72

Therefore, P(D) = 25/72.

Question 14

A vector **v** makes angles α, β, and γ with the positive x, y, and z axes respectively. If cos α = 1/2 and cos β = 1/3, then cos γ is equal to:

Accepted Answer

The direction cosines of a vector **v** are cos α, cos β, and cos γ, where α, β, γ are the angles with the positive x, y, z axes respectively. These satisfy the fundamental identity:

cos² α + cos² β + cos² γ = 1

Step 1: Substitute the given values:
(1/2)² + (1/3)² + cos² γ = 1

Step 2: Compute the squares:
1/4 + 1/9 + cos² γ = 1

Step 3: Find a common denominator (LCM of 4 and 9 is 36):
9/36 + 4/36 + cos² γ = 1
13/36 + cos² γ = 1

Step 4: Solve for cos² γ:
cos² γ = 1 − 13/36 = 36/36 − 13/36 = 23/36

Step 5: Take the square root:
cos γ = ±√(23/36) = ±√23/6

Since the problem asks for cos γ without specifying the sign, the answer is ±√23/6. However, if the question expects a single answer, the positive value is typically given as √23/6.

Question 15

Two vectors **a** = (1, 2, 3) and **b** = (2, −1, 1) are given. A third vector **c** is perpendicular to both **a** and **b**. If **c** · (1, 1, 1) = 5, then **c** is equal to:

Accepted Answer

Step 1: Since **c** is perpendicular to both **a** and **b**, we have **c** = k(**a** × **b**) for some scalar k.

Step 2: Compute **a** × **b**:
**a** × **b** = |**i**  **j**  **k**|
              |1   2   3|
              |2  −1   1|
= **i**(2·1 − 3·(−1)) − **j**(1·1 − 3·2) + **k**(1·(−1) − 2·2)
= **i**(2 + 3) − **j**(1 − 6) + **k**(−1 − 4)
= 5**i** + 5**j** − 5**k**
= (5, 5, −5)

Step 3: So **c** = k(5, 5, −5) = (5k, 5k, −5k).

Step 4: Apply the condition **c** · (1, 1, 1) = 5:
(5k, 5k, −5k) · (1, 1, 1) = 5k + 5k − 5k = 5k = 5
Therefore k = 1.

Step 5: **c** = (5, 5, −5).

Question 16

Let **u** = (cos θ, sin θ, 0) and **v** = (sin θ, −cos θ, 1) be two vectors. The angle α between **u** and **v** satisfies cos α = k/√2, where k is a constant. The value of k is:

Accepted Answer

Step 1: Compute the dot product **u** · **v**:
**u** · **v** = (cos θ)(sin θ) + (sin θ)(−cos θ) + (0)(1)
= cos θ sin θ − sin θ cos θ + 0
= 0

Step 2: Compute the magnitudes:
|**u**| = √(cos² θ + sin² θ + 0²) = √1 = 1
|**v**| = √(sin² θ + cos² θ + 1²) = √(1 + 1) = √2

Step 3: Apply the formula for the angle between two vectors:
cos α = (**u** · **v**) / (|**u**| · |**v**|) = 0 / (1 · √2) = 0

Step 4: Compare with the given form cos α = k/√2:
0 = k/√2
Therefore k = 0.

Question 17

A plane passes through the point (1, 2, 3) and is perpendicular to the vector **n** = (2, −1, 2). The distance from the origin O(0, 0, 0) to this plane is:

Accepted Answer

Step 1: Write the equation of the plane. A plane perpendicular to **n** = (2, −1, 2) and passing through (1, 2, 3) has equation:
2(x − 1) − 1(y − 2) + 2(z − 3) = 0
2x − 2 − y + 2 + 2z − 6 = 0
2x − y + 2z − 6 = 0

Step 2: Rewrite in standard form:
2x − y + 2z = 6

Step 3: Apply the distance formula from a point (x₀, y₀, z₀) to the plane ax + by + cz = d:
distance = |ax₀ + by₀ + cz₀ − d| / √(a² + b² + c²)

Step 4: Substitute O(0, 0, 0), a = 2, b = −1, c = 2, d = 6:
distance = |2(0) − 1(0) + 2(0) − 6| / √(2² + (−1)² + 2²)
= |0 − 6| / √(4 + 1 + 4)
= 6 / √9
= 6 / 3
= 2

Question 18

In a random experiment, two events A and B are such that P(A) = 1/3, P(B) = 1/2, and P(A ∩ B) = 1/6. A vector **u** is defined as **u** = (P(A ∪ B), P(A' ∩ B), P(A ∩ B')), where A' and B' denote the complements of A and B. The magnitude of **u** is:

Accepted Answer

Step 1: Compute P(A ∪ B) using the inclusion-exclusion principle:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 1/3 + 1/2 − 1/6
= 2/6 + 3/6 − 1/6 = 4/6 = 2/3

Step 2: Compute P(A' ∩ B). This is the probability that B occurs but A does not:
P(A' ∩ B) = P(B) − P(A ∩ B) = 1/2 − 1/6 = 3/6 − 1/6 = 2/6 = 1/3

Step 3: Compute P(A ∩ B'). This is the probability that A occurs but B does not:
P(A ∩ B') = P(A) − P(A ∩ B) = 1/3 − 1/6 = 2/6 − 1/6 = 1/6

Step 4: Form the vector **u**:
**u** = (2/3, 1/3, 1/6)

Step 5: Compute the magnitude:
|**u**| = √((2/3)² + (1/3)² + (1/6)²)
= √(4/9 + 1/9 + 1/36)
= √(16/36 + 4/36 + 1/36)
= √(21/36)
= √21/6

Question 19

Two vectors **a** = (1, 2, 3) and **b** = (2, −1, 1) are given. A third vector **c** is perpendicular to both **a** and **b**. If **c** · (1, 1, 1) = 5, find **c**.

Accepted Answer

A vector perpendicular to both **a** and **b** is parallel to **a** × **b**.

Step 1: Compute **a** × **b**.
**a** × **b** = |**i**  **j**  **k**|
              |1   2   3|
              |2  −1   1|
= **i**(2·1 − 3·(−1)) − **j**(1·1 − 3·2) + **k**(1·(−1) − 2·2)
= **i**(2 + 3) − **j**(1 − 6) + **k**(−1 − 4)
= 5**i** + 5**j** − 5**k**
= (5, 5, −5)

Step 2: General form of **c** perpendicular to both **a** and **b**:
**c** = λ(5, 5, −5) = (5λ, 5λ, −5λ) for some scalar λ.

Step 3: Apply the condition **c** · (1, 1, 1) = 5.
(5λ, 5λ, −5λ) · (1, 1, 1) = 5λ + 5λ − 5λ = 5λ = 5
Therefore λ = 1.

Step 4: **c** = (5, 5, −5).

Verification:
- **c** · **a** = 5(1) + 5(2) + (−5)(3) = 5 + 10 − 15 = 0 ✓
- **c** · **b** = 5(2) + 5(−1) + (−5)(1) = 10 − 5 − 5 = 0 ✓
- **c** · (1, 1, 1) = 5 + 5 − 5 = 5 ✓

Question 20

A line passes through the point P(1, 2, 3) and is parallel to the vector **d** = (2, −1, 2). The distance from the point Q(4, 5, 6) to this line is d_Q. Find d_Q.

Accepted Answer

Step 1: Write the parametric equation of the line.
The line through P(1, 2, 3) parallel to **d** = (2, −1, 2) is:
**r**(t) = (1, 2, 3) + t(2, −1, 2) = (1 + 2t, 2 − t, 3 + 2t).

Step 2: Find the vector from P to Q.
**PQ** = Q − P = (4, 5, 6) − (1, 2, 3) = (3, 3, 3).

Step 3: Compute the cross product **PQ** × **d**.
**PQ** × **d** = |**i**  **j**  **k**|
              |3   3   3|
              |2  −1   2|
= **i**(3·2 − 3·(−1)) − **j**(3·2 − 3·2) + **k**(3·(−1) − 3·2)
= **i**(6 + 3) − **j**(6 − 6) + **k**(−3 − 6)
= 9**i** + 0**j** − 9**k**
= (9, 0, −9).

Step 4: Compute the magnitude of **PQ** × **d**.
|**PQ** × **d**| = √(9² + 0² + (−9)²) = √(81 + 0 + 81) = √162 = 9√2.

Step 5: Compute the magnitude of **d**.
|**d**| = √(2² + (−1)² + 2²) = √(4 + 1 + 4) = √9 = 3.

Step 6: Apply the distance formula.
The distance from Q to the line is:
d_Q = |**PQ** × **d**| / |**d**| = 9√2 / 3 = 3√2.

Verification: The formula d = |(**PQ**) × **d**| / |**d**| is the standard formula for point-to-line distance in 3D, derived from the fact that the area of the parallelogram formed by **PQ** and **d** is |**PQ** × **d**|, and dividing by |**d**| gives the height (perpendicular distance).

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