Question

3. Given the displacement vectors A=(2i-5j+3k)m and B=(6i+9j-13k)m ,
Find the magnitude and the direction of the vectors .
A, C=A+4B
B, D=2A-3B

Answer 1

To find the magnitude and direction of the vectors \(A\), \(C\), and \(D\), we will follow these steps:

### Given Vectors:
- \( A = 2\mathbf{i} - 5\mathbf{j} + 3\mathbf{k} \, \text{m} \)
- \( B = 6\mathbf{i} + 9\mathbf{j} - 13\mathbf{k} \, \text{m} \)

### Step 1: Calculate the resultant vectors

1. **Vector C: \( C = A + 4B \)**
\[
C = (2\mathbf{i} - 5\mathbf{j} + 3\mathbf{k}) + 4(6\mathbf{i} + 9\mathbf{j} - 13\mathbf{k})
\]
\[
= (2 + 24)\mathbf{i} + (-5 + 36)\mathbf{j} + (3 - 52)\mathbf{k}
\]
\[
= 26\mathbf{i} + 31\mathbf{j} - 49\mathbf{k} \, \text{m}
\]

2. **Vector D: \( D = 2A - 3B \)**
\[
D = 2(2\mathbf{i} - 5\mathbf{j} + 3\mathbf{k}) - 3(6\mathbf{i} + 9\mathbf{j} - 13\mathbf{k})
\]
\[
= (4 - 18)\mathbf{i} + (-10 - 27)\mathbf{j} + (6 + 39)\mathbf{k}
\]
\[
= -14\mathbf{i} - 37\mathbf{j} + 45\mathbf{k} \, \text{m}
\]

### Step 2: Calculate the magnitude and direction of A, C, and D

**Magnitude of a vector \( \mathbf{V} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \)** is calculated as:
\[
|\mathbf{V}| = \sqrt{a^2 + b^2 + c^2}
\]

1. **Magnitude of A**:
\[
|A| = \sqrt{2^2 + (-5)^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38} \approx 6.16 \, \text{m}
\]

2. **Magnitude of C**:
\[
|C| = \sqrt{26^2 + 31^2 + (-49)^2} = \sqrt{676 + 961 + 2401} = \sqrt{4038} \approx 63.6 \, \text{m}
\]

3. **Magnitude of D**:
\[
|D| = \sqrt{(-14)^2 + (-37)^2 + 45^2} = \sqrt{196 + 1369 + 2025} = \sqrt{3590} \approx 59.9 \, \text{m}
\]

### Direction:
The direction can be represented as a unit vector for each vector. The unit vector \( \hat{V} \) of a vector \( \mathbf{V} \) is given by:
\[
\hat{V} = \frac{\mathbf{V}}{|\mathbf{V}|}
\]

1. **Direction of A**:
\[
\hat{A} = \frac{A}{|A|} = \frac{2\mathbf{i} - 5\mathbf{j} + 3\mathbf{k}}{\sqrt{38}} \approx \left( \frac{2}{\sqrt{38}}, \frac{-5}{\sqrt{38}}, \frac{3}{\sqrt{38}} \right)
\]

2. **Direction of C**:
\[
\hat{C} = \frac{C}{|C|} = \frac{26\mathbf{i} + 31\mathbf{j} - 49\mathbf{k}}{\sqrt{4038}} \approx \left( \frac{26}{\sqrt{4038}}, \frac{31}{\sqrt{4038}}, \frac{-49}{\sqrt{4038}} \right)
\]

3. **Direction of D**:
\[
\hat{D} = \frac{D}{|D|} = \frac{-14\mathbf{i} - 37\mathbf{j} + 45\mathbf{k}}{\sqrt{3590}} \approx \left( \frac{-14}{\sqrt{3590}}, \frac{-37}{\sqrt{3590}}, \frac{45}{\sqrt{3590}} \right)
\]

### Summary:
- **Vector A**: Magnitude \(\approx 6.16 \, \text{m}\), Direction \(\left( \frac{2}{\sqrt{38}}, \frac{-5}{\sqrt{38}}, \frac{3}{\sqrt{38}} \right)\)
- **Vector C**: Magnitude \(\approx 63.6 \, \text{m}\), Direction \(\left( \frac{26}{\sqrt{4038}}, \frac{31}{\sqrt{4038}}, \frac{-49}{\sqrt{4038}} \right)\)
- **Vector D**: Magnitude \(\approx 59.9 \, \text{m}\), Direction \(\left( \frac{-14}{\sqrt{3590}}, \frac{-37}{\sqrt{3590}}, \frac{45}{\sqrt{3590}} \right)\)