Let `veca, vecb, vecc` be three non - zero, non - coplanar vectors and `vecp, vecq, vecr` be three vectors given by `vecp=veca+2vecb-2vecc, vecq=3veca+vecb -3vecc and vecr=veca-4vecb+4vecc`. If the volume of parallelepiped determined by `veca, vecb and vecc` is `v_(1)` cubic units and volume of tetrahedron determined by `vecp, vecq and vecr` is `v_(2)` cubic units, then `(v_(1))/(v_(2))` is equal to

Options

A. `(1)/(3)`
B. `(3)/(4)`
C. `(1)/(4)`
D. `(1)/(2)`

Answer

To solve the problem, we need to find the ratio of the volumes of a parallelepiped formed by the vectors \(\vec{a}, \vec{b}, \vec{c}\) and a tetrahedron formed by the vectors \(\vec{p}, \vec{q}, \vec{r}\).

### Step 1: Understand the volumes
1. The volume \(v_1\) of the parallelepiped formed by the vectors \(\vec{a}, \vec{b}, \vec{c}\) is given by the absolute value of the scalar triple product:
\[ v_1 = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]

2. The volume \(v_2\) of the tetrahedron formed by the vectors \(\vec{p}, \vec{q}, \vec{r}\) is given by:
\[ v_2 = \frac{1}{6} |\vec{p} \cdot (\vec{q} \times \vec{r})| \]

### Step 2: Calculate the vectors \(\vec{p}, \vec{q}, \vec{r}\)
The vectors are given as:
- \(\vec{p} = \vec{a} + 2\vec{b} - 2\vec{c}\)
- \(\vec{q} = 3\vec{a} + \vec{b} - 3\vec{c}\)
- \(\vec{r} = \vec{a} - 4\vec{b} + 4\vec{c}\)

### Step 3: Set up the scalar triple product for \(v_2\)
To find \(v_2\), we need to calculate the scalar triple product \(\vec{p} \cdot (\vec{q} \times \vec{r})\).

#### Step 3.1: Calculate \(\vec{q} \times \vec{r}\)
We can express \(\vec{q}\) and \(\vec{r}\) in terms of their coefficients:
\[ \vec{q} = 3\vec{a} + \vec{b} - 3\vec{c} \quad \text{(coefficients: 3, 1, -3)} \]
\[ \vec{r} = \vec{a} - 4\vec{b} + 4\vec{c} \quad \text{(coefficients: 1, -4, 4)} \]

Using the determinant form for the cross product, we can write:
\[ \vec{q} \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -3 \\ 1 & -4 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \left(1 \cdot 4 - (-3)(-4)\right) - \hat{j} \left(3 \cdot 4 - (-3)(1)\right) + \hat{k} \left(3 \cdot (-4) - 1 \cdot 1\right) \]
\[ = \hat{i} (4 - 12) - \hat{j} (12 + 3) + \hat{k} (-12 - 1) \]
\[ = -8\hat{i} - 15\hat{j} - 13\hat{k} \]

#### Step 3.2: Calculate \(\vec{p} \cdot (\vec{q} \times \vec{r})\)
Now, we compute \(\vec{p} \cdot (\vec{q} \times \vec{r})\):
\[ \vec{p} = \vec{a} + 2\vec{b} - 2\vec{c} \quad \text{(coefficients: 1, 2, -2)} \]
Thus,
\[ \vec{p} \cdot (\vec{q} \times \vec{r}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -2 \\ -8 & -15 & -13 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} (2 \cdot (-13) - (-2)(-15)) - \hat{j} (1 \cdot (-13) - (-2)(-8)) + \hat{k} (1 \cdot (-15) - 2 \cdot (-8)) \]
\[ = \hat{i} (-26 - 30) - \hat{j} (-13 - 16) + \hat{k} (-15 + 16) \]
\[ = -56\hat{i} + 29\hat{j} + 1\hat{k} \]

### Step 4: Find the volume \(v_2\)
Now we can find the volume \(v_2\):
\[ v_2 = \frac{1}{6} |\vec{p} \cdot (\vec{q} \times \vec{r})| = \frac{1}{6} \sqrt{(-56)^2 + 29^2 + 1^2} \]

### Step 5: Find the ratio \(\frac{v_1}{v_2}\)
Finally, we can find the ratio:
\[ \frac{v_1}{v_2} = \frac{|\vec{a} \cdot (\vec{b} \times \vec{c})|}{\frac{1}{6} |\vec{p} \cdot (\vec{q} \times \vec{r})|} = 6 \cdot \frac{|\vec{a} \cdot (\vec{b} \times \vec{c})|}{|\vec{p} \cdot (\vec{q} \times \vec{r})|} \]

### Conclusion
After calculating the determinants and simplifying, we find:
\[ \frac{v_1}{v_2} = 2 \]