Let A, B, C be points with position vectors `r_(1)=2hati-hatj+hatk, r_(2)=hati+2hatj+3hatk` and `r_(3)=3hati+hatj+2hatk` relative to the origin 'O'. The shortest distance between point B and plane OAC is

Options

A. 10
B. 5
C. `sqrt(5//7)`
D. `2sqrt(5//7)`

Answer

To find the shortest distance between point B and the plane OAC, we will follow these steps:

### Step 1: Identify the position vectors
Given the position vectors:
- \( \mathbf{r_1} = 2\hat{i} - \hat{j} + \hat{k} \) (Point A)
- \( \mathbf{r_2} = \hat{i} + 2\hat{j} + 3\hat{k} \) (Point B)
- \( \mathbf{r_3} = 3\hat{i} + \hat{j} + 2\hat{k} \) (Point C)

### Step 2: Find the vectors OA and OC
The vectors from the origin O to points A and C are:
- \( \mathbf{OA} = \mathbf{r_1} = 2\hat{i} - \hat{j} + \hat{k} \)
- \( \mathbf{OC} = \mathbf{r_3} = 3\hat{i} + \hat{j} + 2\hat{k} \)

### Step 3: Calculate the cross product \( \mathbf{OA} \times \mathbf{OC} \)
To find the normal vector to the plane OAC, we calculate the cross product:
\[ \mathbf{OA} \times \mathbf{OC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 1 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \left((-1) \cdot 2 - 1 \cdot 1\right) - \hat{j} \left(2 \cdot 2 - 1 \cdot 3\right) + \hat{k} \left(2 \cdot 1 - (-1) \cdot 3\right) \]
\[ = \hat{i} (-2 - 1) - \hat{j} (4 - 3) + \hat{k} (2 + 3) \]
\[ = -3\hat{i} - 1\hat{j} + 5\hat{k} \]
Thus, \( \mathbf{n} = -3\hat{i} - \hat{j} + 5\hat{k} \).

### Step 4: Find the magnitude of the normal vector
The magnitude of the normal vector \( \mathbf{n} \) is:
\[ |\mathbf{n}| = \sqrt{(-3)^2 + (-1)^2 + 5^2} = \sqrt{9 + 1 + 25} = \sqrt{35} \]

### Step 5: Calculate the distance from point B to the plane OAC
The formula for the distance \( d \) from a point \( \mathbf{P} \) to a plane defined by the normal vector \( \mathbf{n} \) and a point \( \mathbf{A} \) on the plane is:
\[ d = \frac{|\mathbf{n} \cdot (\mathbf{P} - \mathbf{A})|}{|\mathbf{n}|} \]
Here, \( \mathbf{P} = \mathbf{r_2} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \mathbf{A} = \mathbf{r_1} = 2\hat{i} - \hat{j} + \hat{k} \).

Calculating \( \mathbf{P} - \mathbf{A} \):
\[ \mathbf{P} - \mathbf{A} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} + 3\hat{j} + 2\hat{k} \]

Now calculate the dot product \( \mathbf{n} \cdot (\mathbf{P} - \mathbf{A}) \):
\[ \mathbf{n} \cdot (\mathbf{P} - \mathbf{A}) = (-3\hat{i} - \hat{j} + 5\hat{k}) \cdot (-\hat{i} + 3\hat{j} + 2\hat{k}) \]
\[ = (-3)(-1) + (-1)(3) + (5)(2) = 3 - 3 + 10 = 10 \]

### Step 6: Substitute into the distance formula
Now substituting into the distance formula:
\[ d = \frac{|10|}{\sqrt{35}} = \frac{10}{\sqrt{35}} = \frac{10\sqrt{35}}{35} = \frac{2\sqrt{5}}{7} \]

### Final Answer
The shortest distance between point B and plane OAC is:
\[ \frac{2\sqrt{5}}{7} \]