If `atimesb=c, btimesc=a, ctimesa=b`. If vectors a, b and c are forming a right handed system, then the volume of tetrahedron formed by vectors `3a-2b+2c, -a-2c and 2a-3b+4c` is

Answer

To find the volume of the tetrahedron formed by the vectors \( \mathbf{v_1} = 3\mathbf{a} - 2\mathbf{b} + 2\mathbf{c} \), \( \mathbf{v_2} = -\mathbf{a} - 2\mathbf{c} \), and \( \mathbf{v_3} = 2\mathbf{a} - 3\mathbf{b} + 4\mathbf{c} \), we will use the formula for the volume \( V \) of a tetrahedron given by:

\[ V = \frac{1}{6} | \mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3}) | \]

### Step 1: Calculate the cross product \( \mathbf{v_2} \times \mathbf{v_3} \)

First, we need to compute the cross product \( \mathbf{v_2} \times \mathbf{v_3} \).

\[ \mathbf{v_2} = -\mathbf{a} - 2\mathbf{c} \]
\[ \mathbf{v_3} = 2\mathbf{a} - 3\mathbf{b} + 4\mathbf{c} \]

Using the distributive property of the cross product:

\[ \mathbf{v_2} \times \mathbf{v_3} = (-\mathbf{a} - 2\mathbf{c}) \times (2\mathbf{a} - 3\mathbf{b} + 4\mathbf{c}) \]

Expanding this using the distributive property:

\[ = -\mathbf{a} \times (2\mathbf{a}) + 3\mathbf{a} \times \mathbf{b} - 4\mathbf{a} \times \mathbf{c} - 2\mathbf{c} \times (2\mathbf{a}) + 6\mathbf{c} \times \mathbf{b} - 8\mathbf{c} \times \mathbf{c} \]

Since \( \mathbf{a} \times \mathbf{a} = \mathbf{0} \) and \( \mathbf{c} \times \mathbf{c} = \mathbf{0} \):

\[ = 3\mathbf{a} \times \mathbf{b} - 4\mathbf{a} \times \mathbf{c} - 4\mathbf{c} \times \mathbf{a} + 6\mathbf{c} \times \mathbf{b} \]

Using the property \( \mathbf{c} \times \mathbf{a} = -\mathbf{a} \times \mathbf{c} \):

\[ = 3\mathbf{a} \times \mathbf{b} - 4\mathbf{a} \times \mathbf{c} + 4\mathbf{a} \times \mathbf{c} + 6\mathbf{c} \times \mathbf{b} \]

This simplifies to:

\[ = 3\mathbf{a} \times \mathbf{b} + 6\mathbf{c} \times \mathbf{b} \]

### Step 2: Calculate the dot product \( \mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3}) \)

Now we compute \( \mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3}) \):

\[ \mathbf{v_1} = 3\mathbf{a} - 2\mathbf{b} + 2\mathbf{c} \]

So,

\[ \mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3}) = (3\mathbf{a} - 2\mathbf{b} + 2\mathbf{c}) \cdot (3\mathbf{a} \times \mathbf{b} + 6\mathbf{c} \times \mathbf{b}) \]

Using the distributive property:

\[ = 3\mathbf{a} \cdot (3\mathbf{a} \times \mathbf{b}) - 2\mathbf{b} \cdot (3\mathbf{a} \times \mathbf{b}) + 2\mathbf{c} \cdot (3\mathbf{a} \times \mathbf{b}) + 3\mathbf{a} \cdot (6\mathbf{c} \times \mathbf{b}) - 2\mathbf{b} \cdot (6\mathbf{c} \times \mathbf{b}) + 2\mathbf{c} \cdot (6\mathbf{c} \times \mathbf{b}) \]

Using the property \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \mathbf{v} \cdot (\mathbf{w} \times \mathbf{u}) \):

\[ = 0 - 0 + 2\mathbf{c} \cdot (3\mathbf{a} \times \mathbf{b}) + 6(0) - 0 + 0 \]

Thus, we have:

\[ = 6 \cdot \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) \]

### Step 3: Calculate the volume \( V \)

Now we can find the volume:

\[ V = \frac{1}{6} |6 \cdot \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})| \]

Since \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \):

\[ = \frac{1}{6} |6| = 1 \]

Thus, the volume of the tetrahedron is:

\[ \boxed{1} \]