A line `L_1` passing through a point with position vector `p=i+2h+3k` and parallel `a=i+2j+3k`, Another line `L_2` passing through a point with direction vector to `b=3i+j+2k`. Q. The minimum distance of origin from the plane passing through the point with position vector p and perpendicular to the line `L_2`, is

Options

A. (a)`sqrt(14)`
B. (b)`(7)/(sqrt(14))`
C. (c)`(11)/(sqrt(14))`
D. (d)None of these

Answer

To solve the problem, we need to find the minimum distance from the origin to the plane that passes through the point with position vector \( \mathbf{p} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) and is perpendicular to the line \( L_2 \) with direction vector \( \mathbf{b} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k} \).

### Step 1: Identify the normal vector of the plane
The normal vector \( \mathbf{n} \) to the plane is the same as the direction vector of line \( L_2 \):
\[ \mathbf{n} = \mathbf{b} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k} \]

### Step 2: Write the equation of the plane
The general equation of a plane can be expressed as:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( (n_1, n_2, n_3) \) is the normal vector.

Using the point \( \mathbf{p} = (1, 2, 3) \) and the normal vector \( \mathbf{n} = (3, 1, 2) \), we can substitute into the equation:
\[ 3(x - 1) + 1(y - 2) + 2(z - 3) = 0 \]

### Step 3: Simplify the equation of the plane
Expanding the equation:
\[ 3x - 3 + y - 2 + 2z - 6 = 0 \]
Combining like terms gives:
\[ 3x + y + 2z - 11 = 0 \]
Thus, the equation of the plane is:
\[ 3x + y + 2z = 11 \]

### Step 4: Calculate the minimum distance from the origin to the plane
The formula for the distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ D = \frac{|Ax_0 + By_0 + Cz_0 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
In our case, \( A = 3, B = 1, C = 2, D = 11 \), and the point is the origin \( (0, 0, 0) \).

Substituting these values into the formula:
\[ D = \frac{|3(0) + 1(0) + 2(0) - 11|}{\sqrt{3^2 + 1^2 + 2^2}} = \frac{|0 - 11|}{\sqrt{9 + 1 + 4}} = \frac{11}{\sqrt{14}} \]

### Step 5: Final answer
Thus, the minimum distance from the origin to the plane is:
\[ D = \frac{11}{\sqrt{14}} \]