Find the angles which the following vectors, makes form the co-ordinates axes :
(i) `2hati+hatj+3hatk` , (ii) `3hati-4hatj+5hatk`

Answer

To find the angles that the given vectors make with the coordinate axes, we can use the following steps:

### Step 1: Define the Vectors
Let the first vector be \( \mathbf{A} = 2\hat{i} + \hat{j} + 3\hat{k} \) and the second vector be \( \mathbf{B} = 3\hat{i} - 4\hat{j} + 5\hat{k} \).

### Step 2: Calculate the Magnitude of the Vectors
For vector \( \mathbf{A} \):
\[ |\mathbf{A}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]

For vector \( \mathbf{B} \):
\[ |\mathbf{B}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 3: Find the Angles with the Coordinate Axes
The angles \( \alpha \), \( \beta \), and \( \gamma \) that the vector makes with the x-axis, y-axis, and z-axis respectively can be found using the cosine formula:
\[ \cos(\theta) = \frac{\text{component along the axis}}{\text{magnitude of the vector}} \]

#### For Vector \( \mathbf{A} \):
1. **Angle \( \alpha \) with x-axis**:
\[ \cos(\alpha) = \frac{2}{|\mathbf{A}|} = \frac{2}{\sqrt{14}} \]
\[ \alpha = \cos^{-1}\left(\frac{2}{\sqrt{14}}\right) \]

2. **Angle \( \beta \) with y-axis**:
\[ \cos(\beta) = \frac{1}{|\mathbf{A}|} = \frac{1}{\sqrt{14}} \]
\[ \beta = \cos^{-1}\left(\frac{1}{\sqrt{14}}\right) \]

3. **Angle \( \gamma \) with z-axis**:
\[ \cos(\gamma) = \frac{3}{|\mathbf{A}|} = \frac{3}{\sqrt{14}} \]
\[ \gamma = \cos^{-1}\left(\frac{3}{\sqrt{14}}\right) \]

#### For Vector \( \mathbf{B} \):
1. **Angle \( \alpha \) with x-axis**:
\[ \cos(\alpha) = \frac{3}{|\mathbf{B}|} = \frac{3}{5\sqrt{2}} \]
\[ \alpha = \cos^{-1}\left(\frac{3}{5\sqrt{2}}\right) \]

2. **Angle \( \beta \) with y-axis**:
\[ \cos(\beta) = \frac{-4}{|\mathbf{B}|} = \frac{-4}{5\sqrt{2}} \]
\[ \beta = \cos^{-1}\left(\frac{-4}{5\sqrt{2}}\right) \]

3. **Angle \( \gamma \) with z-axis**:
\[ \cos(\gamma) = \frac{5}{|\mathbf{B}|} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \]
\[ \gamma = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \]

### Summary of Results
For vector \( \mathbf{A} \):
- \( \alpha = \cos^{-1}\left(\frac{2}{\sqrt{14}}\right) \)
- \( \beta = \cos^{-1}\left(\frac{1}{\sqrt{14}}\right) \)
- \( \gamma = \cos^{-1}\left(\frac{3}{\sqrt{14}}\right) \)

For vector \( \mathbf{B} \):
- \( \alpha = \cos^{-1}\left(\frac{3}{5\sqrt{2}}\right) \)
- \( \beta = \cos^{-1}\left(\frac{-4}{5\sqrt{2}}\right) \)
- \( \gamma = \frac{\pi}{4} \)