If `vec(a)=7hat(i)-2hat(j)+3hat(k), vec(b)=hat(i)-hat(j)+2hat(k), vec(c )=2hat(i)+8hat(j)`, then find `vec(a).(vec(b)xx vec(c ))` and `(vec(b)xx vec(c )).vec(a)`.

Answer

To solve the problem, we need to calculate two expressions: \(\vec{a} \cdot (\vec{b} \times \vec{c})\) and \((\vec{b} \times \vec{c}) \cdot \vec{a}\).

Given:
\[ \vec{a} = 7\hat{i} - 2\hat{j} + 3\hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \]
\[ \vec{c} = 2\hat{i} + 8\hat{j} \]

### Step 1: Calculate \(\vec{b} \times \vec{c}\)

To find \(\vec{b} \times \vec{c}\), we can use the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of \(\vec{b}\) and \(\vec{c}\):

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 8 & 0 \end{vmatrix} \]

Calculating the determinant, we expand it as follows:

\[ \vec{b} \times \vec{c} = \hat{i} \begin{vmatrix} -1 & 2 \\ 8 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 2 & 8 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -1 & 2 \\ 8 & 0 \end{vmatrix} = (-1)(0) - (2)(8) = -16\)
2. \(\begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} = (1)(0) - (2)(2) = -4\)
3. \(\begin{vmatrix} 1 & -1 \\ 2 & 8 \end{vmatrix} = (1)(8) - (-1)(2) = 8 + 2 = 10\)

Putting it all together:

\[ \vec{b} \times \vec{c} = -16\hat{i} + 4\hat{j} + 10\hat{k} \]

### Step 2: Calculate \(\vec{a} \cdot (\vec{b} \times \vec{c})\)

Now we find \(\vec{a} \cdot (\vec{b} \times \vec{c})\):

\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (7\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (-16\hat{i} + 4\hat{j} + 10\hat{k}) \]

Calculating the dot product:

\[ = 7 \cdot (-16) + (-2) \cdot 4 + 3 \cdot 10 \]
\[ = -112 - 8 + 30 \]
\[ = -90 \]

### Step 3: Calculate \((\vec{b} \times \vec{c}) \cdot \vec{a}\)

Since the dot product is commutative, we have:

\[ (\vec{b} \times \vec{c}) \cdot \vec{a} = \vec{a} \cdot (\vec{b} \times \vec{c}) = -90 \]

### Final Answers:

1. \(\vec{a} \cdot (\vec{b} \times \vec{c}) = -90\)
2. \((\vec{b} \times \vec{c}) \cdot \vec{a} = -90\)