Let A be a point on the line `vecr = (1- 5 mu) hati + (3mu - 1) hatj + (5 + 7mu )hatk` and B (8, 2, 6) be a point in the space. Then the value of `mu` for which the vector `vec(AB)` is parallel to the plane x - 4y + 3z = 1 is:

Options

A. `1/4`
B. `- 5/4`
C. `- 1/2`
D. `- 3/4`

Explanation

To solve the problem, we need to find the value of \( \mu \) for which the vector \( \vec{AB} \) is parallel to the plane defined by the equation \( x - 4y + 3z = 1 \).

### Step 1: Identify the points A and B
The point A lies on the line given by the vector equation:
\[ \vec{r} = (1 - 5\mu) \hat{i} + (3\mu - 1) \hat{j} + (5 + 7\mu) \hat{k} \]
The point B is given as \( B(8, 2, 6) \).

### Step 2: Write the coordinates of point A
From the vector equation, we can express the coordinates of point A as:
\[ A(1 - 5\mu, 3\mu - 1, 5 + 7\mu) \]

### Step 3: Find the vector \( \vec{AB} \)
The vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = B - A = (8 - (1 - 5\mu)) \hat{i} + (2 - (3\mu - 1)) \hat{j} + (6 - (5 + 7\mu)) \hat{k} \]
Calculating each component:
- For \( \hat{i} \): \( 8 - (1 - 5\mu) = 8 - 1 + 5\mu = 7 + 5\mu \)
- For \( \hat{j} \): \( 2 - (3\mu - 1) = 2 + 1 - 3\mu = 3 - 3\mu \)
- For \( \hat{k} \): \( 6 - (5 + 7\mu) = 6 - 5 - 7\mu = 1 - 7\mu \)

Thus,
\[ \vec{AB} = (7 + 5\mu) \hat{i} + (3 - 3\mu) \hat{j} + (1 - 7\mu) \hat{k} \]

### Step 4: Find the normal vector of the plane
The equation of the plane is given by:
\[ x - 4y + 3z = 1 \]
The normal vector \( \vec{n} \) to the plane can be obtained from the coefficients of \( x, y, z \):
\[ \vec{n} = \hat{i} - 4\hat{j} + 3\hat{k} \]

### Step 5: Set up the condition for parallelism
For the vector \( \vec{AB} \) to be parallel to the plane, it must be perpendicular to the normal vector \( \vec{n} \). This means:
\[ \vec{AB} \cdot \vec{n} = 0 \]

Calculating the dot product:
\[ (7 + 5\mu)(1) + (3 - 3\mu)(-4) + (1 - 7\mu)(3) = 0 \]

Expanding this:
\[ (7 + 5\mu) - 4(3 - 3\mu) + 3(1 - 7\mu) = 0 \]
\[ 7 + 5\mu - 12 + 12\mu + 3 - 21\mu = 0 \]
Combining like terms:
\[ (5\mu + 12\mu - 21\mu) + (7 - 12 + 3) = 0 \]
\[ -4\mu - 2 = 0 \]

### Step 6: Solve for \( \mu \)
\[ -4\mu = 2 \implies \mu = -\frac{1}{2} \]

Thus, the value of \( \mu \) for which the vector \( \vec{AB} \) is parallel to the plane is:
\[ \mu = -\frac{1}{2} \]