Let the points P, Q and R have position vectors
`r_(1) = 3i-2j-k`
`r_(2) =i+3j+4k`
and `r_(3) =2i+j-2k`
relative to an origin O.
The distance of P from the plane OQR is

Options

A. 2
B. 3
C. 1
D. 5

Answer

To find the distance of point P from the plane OQR, we will follow these steps:

### Step 1: Identify the position vectors
Given the position vectors:
- \( \mathbf{r_1} = 3\mathbf{i} - 2\mathbf{j} - \mathbf{k} \) (for point P)
- \( \mathbf{r_2} = \mathbf{i} + 3\mathbf{j} + 4\mathbf{k} \) (for point Q)
- \( \mathbf{r_3} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k} \) (for point R)

### Step 2: Find the normal vector to the plane OQR
The normal vector \( \mathbf{n} \) to the plane formed by points O, Q, and R can be found using the cross product of the vectors \( \mathbf{r_2} \) and \( \mathbf{r_3} \).

\[ \mathbf{n} = \mathbf{r_2} \times \mathbf{r_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & 4 \\ 2 & 1 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:

\[ \mathbf{n} = \mathbf{i} \begin{vmatrix} 3 & 4 \\ 1 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 4 \\ 2 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 3 & 4 \\ 1 & -2 \end{vmatrix} = (3)(-2) - (4)(1) = -6 - 4 = -10 \)
2. \( \begin{vmatrix} 1 & 4 \\ 2 & -2 \end{vmatrix} = (1)(-2) - (4)(2) = -2 - 8 = -10 \)
3. \( \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = (1)(1) - (3)(2) = 1 - 6 = -5 \)

Thus, we have:

\[ \mathbf{n} = -10\mathbf{i} + 10\mathbf{j} - 5\mathbf{k} \]

### Step 4: Calculate the distance from point P to the plane OQR
The formula for the distance \( d \) from a point to a plane defined by the normal vector \( \mathbf{n} \) and passing through the origin is given by:

\[ d = \frac{|\mathbf{r_1} \cdot \mathbf{n}|}{|\mathbf{n}|} \]

### Step 5: Calculate \( \mathbf{r_1} \cdot \mathbf{n} \)
Calculating the dot product:

\[ \mathbf{r_1} \cdot \mathbf{n} = (3\mathbf{i} - 2\mathbf{j} - \mathbf{k}) \cdot (-10\mathbf{i} + 10\mathbf{j} - 5\mathbf{k}) \]
\[ = 3(-10) + (-2)(10) + (-1)(-5) = -30 - 20 + 5 = -45 \]

### Step 6: Calculate the magnitude of \( \mathbf{n} \)
Calculating the magnitude of \( \mathbf{n} \):

\[ |\mathbf{n}| = \sqrt{(-10)^2 + (10)^2 + (-5)^2} = \sqrt{100 + 100 + 25} = \sqrt{225} = 15 \]

### Step 7: Calculate the distance
Now substituting back into the distance formula:

\[ d = \frac{|-45|}{15} = \frac{45}{15} = 3 \]

Thus, the distance of point P from the plane OQR is **3**.

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