If `M` denotes the md point of the line segment joining `A(4hati+6hatj-10hatk)` and `B(-hati+2hatj+hatk)`, then the equation, of the plane through` M` and perpendicular to AB is

Options

A. `vecr.(-5hati-3hatj+11hatk)+135/2=0`
B. `vecr.(3/2hati+7/2hatj-9/2hatk)+135/2=0`
C. `vecr.(4hati+5hatj-10hatk)+4=0`
D. `vecr.(-hati+2hatj+hatk)+4=0`

Explanation

To find the equation of the plane that passes through the midpoint \( M \) of the line segment joining points \( A(4\hat{i} + 6\hat{j} - 10\hat{k}) \) and \( B(-\hat{i} + 2\hat{j} + \hat{k}) \), and is perpendicular to the line segment \( AB \), we will follow these steps:

### Step 1: Find the Midpoint \( M \)

The formula for the midpoint \( M \) of a line segment joining two points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) is given by:

\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]

For points \( A(4, 6, -10) \) and \( B(-1, 2, 1) \):

\[ M = \left( \frac{4 + (-1)}{2}, \frac{6 + 2}{2}, \frac{-10 + 1}{2} \right) \]

Calculating each component:

\[ M_x = \frac{4 - 1}{2} = \frac{3}{2}, \quad M_y = \frac{6 + 2}{2} = 4, \quad M_z = \frac{-10 + 1}{2} = \frac{-9}{2} \]

Thus, the coordinates of \( M \) are:

\[ M\left( \frac{3}{2}, 4, -\frac{9}{2} \right) \]

### Step 2: Find the Direction Vector \( \overrightarrow{AB} \)

The direction vector \( \overrightarrow{AB} \) can be calculated as:

\[ \overrightarrow{AB} = B - A = (-1 - 4, 2 - 6, 1 - (-10)) \]

Calculating each component:

\[ \overrightarrow{AB} = (-5, -4, 11) \]

### Step 3: Write the Normal Vector

The normal vector \( \mathbf{n} \) to the plane is the same as the direction vector \( \overrightarrow{AB} \):

\[ \mathbf{n} = -5\hat{i} - 4\hat{j} + 11\hat{k} \]

### Step 4: Use the Point-Normal Form of the Plane Equation

The equation of a plane in point-normal form is given by:

\[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0 \]

Where \( \mathbf{r_0} \) is the position vector of point \( M \) and \( \mathbf{n} \) is the normal vector.

Let \( \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k} \) and \( \mathbf{r_0} = \frac{3}{2}\hat{i} + 4\hat{j} - \frac{9}{2}\hat{k} \).

Substituting \( \mathbf{n} \) and \( \mathbf{r_0} \):

\[ (-5\hat{i} - 4\hat{j} + 11\hat{k}) \cdot \left( (x - \frac{3}{2})\hat{i} + (y - 4)\hat{j} + (z + \frac{9}{2})\hat{k} \right) = 0 \]

Calculating the dot product:

\[ -5(x - \frac{3}{2}) - 4(y - 4) + 11(z + \frac{9}{2}) = 0 \]

Expanding this:

\[ -5x + \frac{15}{2} - 4y + 16 + 11z + \frac{99}{2} = 0 \]

Combining the constant terms:

\[ -5x - 4y + 11z + \left( \frac{15 + 32 + 99}{2} \right) = 0 \]

Calculating the constant:

\[ \frac{146}{2} = 73 \]

Thus, the equation of the plane is:

\[ -5x - 4y + 11z = -73 \]

### Final Equation of the Plane

Rearranging gives us:

\[ 5x + 4y - 11z = 73 \]