Let `vecOB = hati + 2hatj + 2hatk " and" vecOA = 4hati + 2hatj + 2hatk` . The distance of the point B from the straight line passing through A and parallel to the vector `2hati + 3hatj + 6hatk` is

Options

A. `(7sqrt5)/(9)`
B. `(5sqrt7)/(9)`
C. `(3sqrt5)/(7)`
D. `(9 sqrt5)/(7)`

Answer

To find the distance of point B from the straight line passing through point A and parallel to the vector \( \vec{d} = 2\hat{i} + 3\hat{j} + 6\hat{k} \), we can follow these steps:

### Step 1: Identify the Points and the Direction Vector
We have:
- Point \( A \) represented by vector \( \vec{OA} = 4\hat{i} + 2\hat{j} + 2\hat{k} \)
- Point \( B \) represented by vector \( \vec{OB} = \hat{i} + 2\hat{j} + 2\hat{k} \)
- The direction vector of the line is \( \vec{d} = 2\hat{i} + 3\hat{j} + 6\hat{k} \)

### Step 2: Write the Equation of the Line
The equation of the line passing through point A and parallel to vector \( \vec{d} \) can be expressed as:
\[ \vec{R} = \vec{OA} + \lambda \vec{d} \]
Substituting the values:
\[ \vec{R} = (4\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \]
This simplifies to:
\[ \vec{R} = (4 + 2\lambda)\hat{i} + (2 + 3\lambda)\hat{j} + (2 + 6\lambda)\hat{k} \]

### Step 3: Find the Vector from Point B to a Point on the Line
Let \( D \) be a point on the line represented by \( \vec{R} \). The vector \( \vec{BD} \) from point B to point D is:
\[ \vec{BD} = \vec{R} - \vec{OB} = [(4 + 2\lambda) - 1]\hat{i} + [(2 + 3\lambda) - 2]\hat{j} + [(2 + 6\lambda) - 2]\hat{k} \]
This simplifies to:
\[ \vec{BD} = (3 + 2\lambda)\hat{i} + (3\lambda)\hat{j} + (6\lambda)\hat{k} \]

### Step 4: Use the Perpendicularity Condition
The distance from point B to the line is the length of the perpendicular dropped from point B to the line. For this, \( \vec{BD} \) must be perpendicular to the direction vector \( \vec{d} \). Thus, we set their dot product to zero:
\[ \vec{BD} \cdot \vec{d} = 0 \]
Calculating the dot product:
\[ (3 + 2\lambda)(2) + (3\lambda)(3) + (6\lambda)(6) = 0 \]
Expanding this gives:
\[ 6 + 4\lambda + 9\lambda + 36\lambda = 0 \]
Combining like terms:
\[ 6 + 49\lambda = 0 \implies \lambda = -\frac{6}{49} \]

### Step 5: Find the Coordinates of Point D
Substituting \( \lambda \) back into the equation for \( \vec{R} \):
\[ \vec{R} = (4 + 2(-\frac{6}{49}))\hat{i} + (2 + 3(-\frac{6}{49}))\hat{j} + (2 + 6(-\frac{6}{49}))\hat{k} \]
Calculating each component:
\[ x_D = 4 - \frac{12}{49} = \frac{196 - 12}{49} = \frac{184}{49} \]
\[ y_D = 2 - \frac{18}{49} = \frac{98 - 18}{49} = \frac{80}{49} \]
\[ z_D = 2 - \frac{36}{49} = \frac{98 - 36}{49} = \frac{62}{49} \]

### Step 6: Calculate the Distance \( BD \)
Now, we can find the distance \( d \) from point B to point D:
\[ d = |\vec{BD}| = \sqrt{(x_D - 1)^2 + (y_D - 2)^2 + (z_D - 2)^2} \]
Calculating each term:
\[ x_D - 1 = \frac{184}{49} - \frac{49}{49} = \frac{135}{49} \]
\[ y_D - 2 = \frac{80}{49} - \frac{98}{49} = -\frac{18}{49} \]
\[ z_D - 2 = \frac{62}{49} - \frac{98}{49} = -\frac{36}{49} \]
Thus,
\[ d = \sqrt{\left(\frac{135}{49}\right)^2 + \left(-\frac{18}{49}\right)^2 + \left(-\frac{36}{49}\right)^2} \]
Calculating:
\[ d = \sqrt{\frac{18225 + 324 + 1296}{2401}} = \sqrt{\frac{19845}{2401}} = \frac{9\sqrt{5}}{7} \]

### Final Answer
The distance of point B from the line is:
\[ \frac{9\sqrt{5}}{7} \]