Let `vecb=-veci+4vecj+6veck, vecc=2veci-7vecj-10veck.` If `veca` be a unit vector and the scalar triple product `[veca vecb vecc]` has the greatest value then `veca` is

A. `(1)/(sqrt(3))(hati+hatj+hatk)`
B. `(1)/(sqrt(5))(sqrt(2)hati-hatj-sqrt(2)hatk)`
C. `(1)/(3)(2hati+2hatj-hatk)`
D. `(1)/(sqrt(59))(3hati-7hatj-hatk)`

Options

A. `(1)/(sqrt(3))(hati+hatj+hatk)`
B. `(1)/(sqrt(5))(sqrt(2)hati-hatj-sqrt(2)hatk)`
C. `(1)/(3)(2hati+2hatj-hatk)`
D. `(1)/(sqrt(59))(3hati-7hatj-hatk)`

Answer

To solve the problem, we need to find the unit vector \(\vec{a}\) such that the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) is maximized. The vectors \(\vec{b}\) and \(\vec{c}\) are given as:

\[ \vec{b} = -\hat{i} + 4\hat{j} + 6\hat{k} \]
\[ \vec{c} = 2\hat{i} - 7\hat{j} - 10\hat{k} \]

### Step 1: Compute the cross product \(\vec{b} \times \vec{c}\)

The cross product \(\vec{b} \times \vec{c}\) can be computed using the determinant of a matrix:

\[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 6 \\ 2 & -7 & -10 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} 4 & 6 \\ -7 & -10 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 6 \\ 2 & -10 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 4 \\ 2 & -7 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 4 & 6 \\ -7 & -10 \end{vmatrix} = (4)(-10) - (6)(-7) = -40 + 42 = 2\)
2. \(\begin{vmatrix} -1 & 6 \\ 2 & -10 \end{vmatrix} = (-1)(-10) - (6)(2) = 10 - 12 = -2\)
3. \(\begin{vmatrix} -1 & 4 \\ 2 & -7 \end{vmatrix} = (-1)(-7) - (4)(2) = 7 - 8 = -1\)

Putting it all together:

\[ \vec{b} \times \vec{c} = 2\hat{i} + 2\hat{j} - \hat{k} \]

### Step 2: Find the unit vector \(\vec{a}\)

To maximize the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\), \(\vec{a}\) must be parallel to \(\vec{b} \times \vec{c}\). Therefore, we can express \(\vec{a}\) as:

\[ \vec{a} = k(2\hat{i} + 2\hat{j} - \hat{k}) \]

where \(k\) is a scalar.

### Step 3: Normalize \(\vec{a}\)

Since \(\vec{a}\) is a unit vector, we need to find \(k\) such that:

\[ |\vec{a}| = 1 \implies |k| \cdot |(2\hat{i} + 2\hat{j} - \hat{k})| = 1 \]

Calculating the magnitude of \((2\hat{i} + 2\hat{j} - \hat{k})\):

\[ |(2\hat{i} + 2\hat{j} - \hat{k})| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

Thus,

\[ |k| \cdot 3 = 1 \implies |k| = \frac{1}{3} \]

### Step 4: Determine the possible values of \(\vec{a}\)

This gives us two possible unit vectors for \(\vec{a}\):

\[ \vec{a} = \frac{1}{3}(2\hat{i} + 2\hat{j} - \hat{k}) \quad \text{or} \quad \vec{a} = -\frac{1}{3}(2\hat{i} + 2\hat{j} - \hat{k}) \]

However, we are looking for the positive scalar, so:

\[ \vec{a} = \frac{1}{3}(2\hat{i} + 2\hat{j} - \hat{k}) \]

### Conclusion

Thus, the correct answer is:

\[ \vec{a} = \frac{1}{3}(2\hat{i} + 2\hat{j} - \hat{k}) \]

This corresponds to option C.