(i) show that the line :
`vec(r) = 2 hati - 3 hatj + 5 hatk + lambda (hati - hatj + 2 hatk)`
lies in the plane `vec(r) . (3 hat(i) + hatj - hatk ) + 2 = 0 `.
(ii) Show that the line :
`vec(r) = hati + hatj + lambda (2 hati + hatj + 4 hatk)`
lies in the plane `vec(r). (hati + 2 hatj - hatk ) = 3.`

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (i)

**Given:**
The line is represented as:
\[ \vec{r} = 2\hat{i} - 3\hat{j} + 5\hat{k} + \lambda(\hat{i} - \hat{j} + 2\hat{k}) \]
The plane is defined by the equation:
\[ \vec{r} \cdot (3\hat{i} + \hat{j} - \hat{k}) + 2 = 0 \]

**Step 1: Identify the fixed point and direction vector of the line.**
- The fixed point \( A \) of the line is \( \vec{A} = 2\hat{i} - 3\hat{j} + 5\hat{k} \).
- The direction vector \( \vec{b} \) of the line is \( \hat{b} = \hat{i} - \hat{j} + 2\hat{k} \).

**Step 2: Check if the fixed point lies in the plane.**
To check if point \( A \) lies in the plane, substitute \( \vec{A} \) into the plane equation:
\[ \vec{A} \cdot (3\hat{i} + \hat{j} - \hat{k}) + 2 = 0 \]
Calculating the dot product:
\[ \vec{A} \cdot (3\hat{i} + \hat{j} - \hat{k}) = (2)(3) + (-3)(1) + (5)(-1) = 6 - 3 - 5 = -2 \]
Now substituting into the plane equation:
\[ -2 + 2 = 0 \]
Since this holds true, point \( A \) lies on the plane.

**Step 3: Check if the direction vector is perpendicular to the normal vector of the plane.**
The normal vector \( \vec{n} \) of the plane is \( 3\hat{i} + \hat{j} - \hat{k} \). We need to check if:
\[ \vec{b} \cdot \vec{n} = 0 \]
Calculating the dot product:
\[ \vec{b} \cdot \vec{n} = (1)(3) + (-1)(1) + (2)(-1) = 3 - 1 - 2 = 0 \]
Since this holds true, the line lies in the plane.

### Conclusion for Part (i):
The line lies in the given plane.

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### Part (ii)

**Given:**
The line is represented as:
\[ \vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} + \hat{j} + 4\hat{k}) \]
The plane is defined by the equation:
\[ \vec{r} \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 3 \]

**Step 1: Identify the fixed point and direction vector of the line.**
- The fixed point \( B \) of the line is \( \vec{B} = \hat{i} + \hat{j} + 0\hat{k} \).
- The direction vector \( \vec{b} \) of the line is \( \hat{b} = 2\hat{i} + \hat{j} + 4\hat{k} \).

**Step 2: Check if the fixed point lies in the plane.**
To check if point \( B \) lies in the plane, substitute \( \vec{B} \) into the plane equation:
\[ \vec{B} \cdot (\hat{i} + 2\hat{j} - \hat{k}) = 3 \]
Calculating the dot product:
\[ \vec{B} \cdot (\hat{i} + 2\hat{j} - \hat{k}) = (1)(1) + (1)(2) + (0)(-1) = 1 + 2 + 0 = 3 \]
Since this holds true, point \( B \) lies on the plane.

**Step 3: Check if the direction vector is perpendicular to the normal vector of the plane.**
The normal vector \( \vec{n} \) of the plane is \( \hat{i} + 2\hat{j} - \hat{k} \). We need to check if:
\[ \vec{b} \cdot \vec{n} = 0 \]
Calculating the dot product:
\[ \vec{b} \cdot \vec{n} = (2)(1) + (1)(2) + (4)(-1) = 2 + 2 - 4 = 0 \]
Since this holds true, the line lies in the plane.

### Conclusion for Part (ii):
The line lies in the given plane.

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