Equation of the plane through the mid - point of the line segment joining the points P(4,5,-10) and Q (-1,2,1) and perpendicular to PQ is

Options

A. `vecr.(3/2hati+7/2hatj-9/2hatk)=45`
B. `vecr.(-hati+2hatj-hatk)=135/2`
C. `vecr.(5hati+3hatj-11hatk)+135/2=0`
D. `vecr.(5hati+3hatj-11hatk)+135/2`

Answer

To find the equation of the plane through the midpoint of the line segment joining the points \( P(4, 5, -10) \) and \( Q(-1, 2, 1) \) and perpendicular to the line segment \( PQ \), we can follow these steps:

### Step 1: Find the midpoint of the line segment \( PQ \)

The midpoint \( C \) of the line segment joining points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) is given by the formula:

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

Substituting the coordinates of points \( P(4, 5, -10) \) and \( Q(-1, 2, 1) \):

\[ C\left(\frac{4 + (-1)}{2}, \frac{5 + 2}{2}, \frac{-10 + 1}{2}\right) = C\left(\frac{3}{2}, \frac{7}{2}, \frac{-9}{2}\right) \]

### Step 2: Find the direction vector of line segment \( PQ \)

The direction vector \( \vec{PQ} \) from point \( P \) to point \( Q \) is given by:

\[ \vec{PQ} = Q - P = (-1 - 4, 2 - 5, 1 - (-10)) = (-5, -3, 11) \]

### Step 3: Use the point-normal form of the equation of a plane

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

\[ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \]

where \( (x_0, y_0, z_0) \) is a point on the plane (which is the midpoint \( C \)) and \( (n_x, n_y, n_z) \) is the normal vector to the plane (which is the direction vector \( \vec{PQ} \)).

Substituting \( C\left(\frac{3}{2}, \frac{7}{2}, \frac{-9}{2}\right) \) and the normal vector \( (-5, -3, 11) \):

\[ -5\left(x - \frac{3}{2}\right) - 3\left(y - \frac{7}{2}\right) + 11\left(z + \frac{9}{2}\right) = 0 \]

### Step 4: Expand and simplify the equation

Expanding the equation:

\[ -5x + \frac{15}{2} - 3y + \frac{21}{2} + 11z + \frac{99}{2} = 0 \]

Combining the constant terms:

\[ -5x - 3y + 11z + \left(\frac{15 + 21 + 99}{2}\right) = 0 \]

Calculating the constant:

\[ 15 + 21 + 99 = 135 \quad \Rightarrow \quad -5x - 3y + 11z + \frac{135}{2} = 0 \]

Rearranging gives:

\[ 5x + 3y - 11z = \frac{135}{2} \]

### Final Equation of the Plane

Thus, the equation of the plane is:

\[ 5x + 3y - 11z = \frac{135}{2} \]