(i) Find the angle between the line :
` ( 2 hati + 3 hatj + 4 hatk ) + lambda (2 hati + 3 hatj + 4 hatk ) `
and the plane : `vec(r).(hati + hatj + hatk) = 5 `.
(ii) Fiind the angle between the line joining (3,-4,-2)and (12,2,0) and the plane `vec(r). (hati + hatj + hatk) = 4`

Answer

To solve the given problems step by step, we will break down each part of the question.

### Part (i): Find the angle between the line and the plane.

**Step 1: Identify the direction vector of the line.**
The line is given by:
\[ \vec{L} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + \lambda(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \]
The direction vector \( \vec{b} \) of the line is:
\[ \vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \]

**Step 2: Identify the normal vector of the plane.**
The equation of the plane is given by:
\[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 5 \]
The normal vector \( \vec{n} \) of the plane is:
\[ \vec{n} = \hat{i} + \hat{j} + \hat{k} \]

**Step 3: Calculate the angle between the direction vector and the normal vector.**
The angle \( \phi \) between the normal vector \( \vec{n} \) and the direction vector \( \vec{b} \) can be found using the dot product formula:
\[ \cos \phi = \frac{\vec{n} \cdot \vec{b}}{|\vec{n}| |\vec{b}|} \]

**Step 4: Calculate the dot product \( \vec{n} \cdot \vec{b} \).**
\[ \vec{n} \cdot \vec{b} = (1)(2) + (1)(3) + (1)(4) = 2 + 3 + 4 = 9 \]

**Step 5: Calculate the magnitudes of \( \vec{n} \) and \( \vec{b} \).**
\[ |\vec{n}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
\[ |\vec{b}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

**Step 6: Substitute into the cosine formula.**
\[ \cos \phi = \frac{9}{\sqrt{3} \cdot \sqrt{29}} = \frac{9}{\sqrt{87}} \]

**Step 7: Find the angle \( \theta \) between the line and the plane.**
Since \( \theta = 90^\circ - \phi \), we can find \( \sin \theta \):
\[ \sin \theta = \cos \phi = \frac{9}{\sqrt{87}} \]

**Step 8: Calculate \( \theta \).**
\[ \theta = \sin^{-1} \left(\frac{9}{\sqrt{87}}\right) \]

### Part (ii): Find the angle between the line joining (3, -4, -2) and (12, 2, 0) and the plane.

**Step 1: Find the direction vector of the line joining the two points.**
Let the points be \( A(3, -4, -2) \) and \( B(12, 2, 0) \).
The direction vector \( \vec{b} \) is:
\[ \vec{b} = (12 - 3) \hat{i} + (2 + 4) \hat{j} + (0 + 2) \hat{k} = 9 \hat{i} + 6 \hat{j} + 2 \hat{k} \]

**Step 2: Identify the normal vector of the plane.**
The normal vector \( \vec{n} \) of the plane is the same as in part (i):
\[ \vec{n} = \hat{i} + \hat{j} + \hat{k} \]

**Step 3: Calculate the angle between the direction vector and the normal vector.**
Using the dot product formula:
\[ \cos \phi = \frac{\vec{n} \cdot \vec{b}}{|\vec{n}| |\vec{b}|} \]

**Step 4: Calculate the dot product \( \vec{n} \cdot \vec{b} \).**
\[ \vec{n} \cdot \vec{b} = (1)(9) + (1)(6) + (1)(2) = 9 + 6 + 2 = 17 \]

**Step 5: Calculate the magnitude of \( \vec{b} \).**
\[ |\vec{b}| = \sqrt{9^2 + 6^2 + 2^2} = \sqrt{81 + 36 + 4} = \sqrt{121} = 11 \]

**Step 6: Substitute into the cosine formula.**
\[ \cos \phi = \frac{17}{\sqrt{3} \cdot 11} \]

**Step 7: Find the angle \( \theta \) between the line and the plane.**
\[ \sin \theta = \cos \phi = \frac{17}{11 \sqrt{3}} \]

**Step 8: Calculate \( \theta \).**
\[ \theta = \sin^{-1} \left(\frac{17}{11 \sqrt{3}}\right) \]