If the line `vec(r) = (-2 hati + 3 hatj + 4 hatk ) + lambda ( - hatk hati + 2 hatj + hatk )` is parallel to the plane `vec(r). (2 hati + 3 hatj - 4 hatk) + 7 = 0, ` then the value of `k` is :

Options

A. 0
B. 1
C. `-1`
D. `-2`

Answer

To solve the problem, we need to determine the value of \( k \) such that the given line is parallel to the specified plane.

### Step-by-Step Solution:

1. **Identify the Direction Ratios of the Line:**
The line is given by the equation:
\[ \vec{r} = (-2 \hat{i} + 3 \hat{j} + 4 \hat{k}) + \lambda (-\hat{i} + 2 \hat{j} + \hat{k}) \]
The direction ratios of the line can be extracted from the vector multiplying \( \lambda \):
\[ \vec{d_1} = (-1, 2, 1) \]

2. **Identify the Normal Vector of the Plane:**
The equation of the plane is:
\[ \vec{r} \cdot (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) + 7 = 0 \]
The normal vector of the plane can be identified as:
\[ \vec{n} = (2, 3, -4) \]

3. **Condition for Parallelism:**
For the line to be parallel to the plane, the direction ratios of the line must be orthogonal to the normal vector of the plane. This means their dot product must equal zero:
\[ \vec{d_1} \cdot \vec{n} = 0 \]

4. **Calculate the Dot Product:**
Substitute the direction ratios and the normal vector into the dot product:
\[ (-1, 2, 1) \cdot (2, 3, -4) = -1 \cdot 2 + 2 \cdot 3 + 1 \cdot (-4) \]
Simplifying this gives:
\[ -2 + 6 - 4 = 0 \]

5. **Set Up the Equation:**
Now, we need to include \( k \) in the direction ratios. The correct direction ratios of the line should be:
\[ (-k, 2, k) \]
Therefore, we need to set up the equation:
\[ (-k, 2, k) \cdot (2, 3, -4) = 0 \]

6. **Calculate the New Dot Product:**
This gives us:
\[ -k \cdot 2 + 2 \cdot 3 + k \cdot (-4) = 0 \]
Simplifying this:
\[ -2k + 6 - 4k = 0 \]
Combining like terms:
\[ -6k + 6 = 0 \]

7. **Solve for \( k \):**
Rearranging gives:
\[ -6k = -6 \implies k = 1 \]

### Final Answer:
The value of \( k \) is \( 1 \).

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