Let `veca=hati+hatj+hatk,vecb=hati+4hatj-hatk and vec c =hati+hatj+2hatk`. If `vecS` be a unit vector, then the magnitude of the vector `(veca.vecS)(vecbxxvecc)+(vecb.vecS)(veccxxveca)+(vecc.vecS)(vecaxxvecb)` is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the magnitude of the vector given by the expression:

\[ (\vec{a} \cdot \vec{S})(\vec{b} \times \vec{c}) + (\vec{b} \cdot \vec{S})(\vec{c} \times \vec{a}) + (\vec{c} \cdot \vec{S})(\vec{a} \times \vec{b}) \]

where \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\), \(\vec{b} = \hat{i} + 4\hat{j} - \hat{k}\), and \(\vec{c} = \hat{i} + \hat{j} + 2\hat{k}\).

### Step 1: Calculate the Scalar Triple Product
The scalar triple product \(\vec{a} \cdot (\vec{b} \times \vec{c})\) gives the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). We can compute this using the determinant of a matrix formed by the components of the vectors.

\[ \text{Determinant} = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 4 & -1 \\ 1 & 1 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ = 1 \cdot \begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} 4 & -1 \\ 1 & 2 \end{vmatrix} = (4 \cdot 2) - (-1 \cdot 1) = 8 + 1 = 9\)

2. \(\begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} = (1 \cdot 2) - (-1 \cdot 1) = 2 + 1 = 3\)

3. \(\begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} = (1 \cdot 1) - (4 \cdot 1) = 1 - 4 = -3\)

Putting it all together:

\[ = 1 \cdot 9 - 1 \cdot 3 + 1 \cdot (-3) = 9 - 3 - 3 = 3 \]

### Step 2: Express the Given Vector in Terms of Scalar Triple Products
The expression we need to evaluate can be simplified using the scalar triple product:

\[ (\vec{a} \cdot \vec{S})(\vec{b} \times \vec{c}) + (\vec{b} \cdot \vec{S})(\vec{c} \times \vec{a}) + (\vec{c} \cdot \vec{S})(\vec{a} \times \vec{b}) = 3(\vec{S} \cdot \vec{a} \times \vec{b} \times \vec{c}) \]

### Step 3: Calculate the Magnitude
Since \(\vec{S}\) is a unit vector, we have:

\[ \text{Magnitude} = |3(\vec{S} \cdot \vec{a} \times \vec{b} \times \vec{c})| = 3 \cdot |\vec{S}| = 3 \cdot 1 = 3 \]

### Final Result
Thus, the magnitude of the vector is:

\[ \boxed{9} \]