Let `OA=a=1/2(i+j-2k), OC=b=i-2j+k" and "OB=10a+2b`. Let p (in `("unit")^(2)`) be the area of the quadrilateral OABC and q (in `("unit")^(2)`) be the area of the parallelogram with OA and OC as adjacent sides, then p/q is equal to

Options

A. 3
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to find the areas \( p \) and \( q \) as described in the question and then compute the ratio \( \frac{p}{q} \).

### Step 1: Define the vectors
Given:
- \( OA = a = \frac{1}{2}(i + j - 2k) \)
- \( OC = b = i - 2j + k \)
- \( OB = 10a + 2b \)

### Step 2: Calculate \( OB \)
First, we need to calculate \( OB \):
\[ OB = 10a + 2b = 10\left(\frac{1}{2}(i + j - 2k)\right) + 2(i - 2j + k) \]
Calculating \( 10a \):
\[ 10a = 5(i + j - 2k) = 5i + 5j - 10k \]
Calculating \( 2b \):
\[ 2b = 2(i - 2j + k) = 2i - 4j + 2k \]
Now, adding these two results:
\[ OB = (5i + 5j - 10k) + (2i - 4j + 2k) = (5 + 2)i + (5 - 4)j + (-10 + 2)k = 7i + 1j - 8k \]

### Step 3: Calculate \( AB \) and \( BC \)
Next, we find \( AB \) and \( BC \):
\[ AB = OB - OA = (7i + j - 8k) - \left(\frac{1}{2}(i + j - 2k)\right) \]
Calculating \( AB \):
\[ AB = (7 - \frac{1}{2})i + (1 - \frac{1}{2})j + (-8 + 1)k = \frac{13}{2}i + \frac{1}{2}j - 7k \]

For \( BC \):
\[ BC = OC - OB = (i - 2j + k) - (7i + j - 8k) \]
Calculating \( BC \):
\[ BC = (1 - 7)i + (-2 - 1)j + (1 + 8)k = -6i - 3j + 9k \]

### Step 4: Calculate the area \( p \) of quadrilateral \( OABC \)
The area \( p \) can be calculated as:
\[ p = \frac{1}{2} | OA \times AB | + \frac{1}{2} | OC \times BC | \]
Calculating \( OA \times AB \):
\[ OA = \frac{1}{2}(i + j - 2k), \quad AB = \frac{13}{2}i + \frac{1}{2}j - 7k \]
Using the determinant method for the cross product:
\[ OA \times AB = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{2} & \frac{1}{2} & -1 \\ \frac{13}{2} & \frac{1}{2} & -7 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}\left(\frac{1}{2}(-7) - (-1)(\frac{1}{2})\right) - \hat{j}\left(\frac{1}{2}(-7) - (-1)(\frac{13}{2})\right) + \hat{k}\left(\frac{1}{2}(\frac{1}{2}) - \frac{1}{2}(\frac{13}{2})\right) \]
\[ = \hat{i}\left(-\frac{7}{2} + \frac{1}{2}\right) - \hat{j}\left(-\frac{7}{2} + \frac{13}{2}\right) + \hat{k}\left(\frac{1}{4} - \frac{13}{4}\right) \]
\[ = \hat{i}\left(-3\right) - \hat{j}\left(3\right) + \hat{k}\left(-3\right) = -3i - 3j - 3k \]
Thus, \( |OA \times AB| = 3\sqrt{3} \).

Calculating \( OC \times BC \):
\[ OC = i - 2j + k, \quad BC = -6i - 3j + 9k \]
Using the same determinant method:
\[ OC \times BC = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ -6 & -3 & 9 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}((-2)(9) - (1)(-3)) - \hat{j}((1)(9) - (1)(-6)) + \hat{k}((1)(-3) - (-2)(-6)) \]
\[ = \hat{i}(-18 + 3) - \hat{j}(9 + 6) + \hat{k}(-3 - 12) \]
\[ = -15\hat{i} - 15\hat{j} - 15\hat{k} \]
Thus, \( |OC \times BC| = 15\sqrt{3} \).

Now we can find \( p \):
\[ p = \frac{1}{2} \cdot 3\sqrt{3} + \frac{1}{2} \cdot 15\sqrt{3} = \frac{18\sqrt{3}}{2} = 9\sqrt{3} \]

### Step 5: Calculate the area \( q \) of the parallelogram
The area \( q \) is given by:
\[ q = |OA \times OC| \]
Calculating \( OA \times OC \):
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{2} & \frac{1}{2} & -1 \\ 1 & -2 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}\left(\frac{1}{2}(1) - (-1)(-2)\right) - \hat{j}\left(\frac{1}{2}(1) - (-1)(1)\right) + \hat{k}\left(\frac{1}{2}(-2) - \frac{1}{2}(1)\right) \]
\[ = \hat{i}\left(\frac{1}{2} - 2\right) - \hat{j}\left(\frac{1}{2} + 1\right) + \hat{k}\left(-1 - \frac{1}{2}\right) \]
\[ = -\frac{3}{2}\hat{i} - \frac{3}{2}\hat{j} - \frac{3}{2}\hat{k} \]
Thus, \( |OA \times OC| = \frac{3\sqrt{3}}{2} \).

### Step 6: Calculate the ratio \( \frac{p}{q} \)
Now we can find the ratio:
\[ \frac{p}{q} = \frac{9\sqrt{3}}{\frac{3\sqrt{3}}{2}} = 9 \cdot \frac{2}{3} = 6 \]

### Final Answer
Thus, the ratio \( \frac{p}{q} \) is \( 6 \).