Let `barb = -1hati+4hatj+6hatk and barc=2hati-7hatj-10hatk` IF `bara` be a unit vector and the scalar triple product `[bara barb barc]` has the greatest value then `bara` is equal to

Options

A. `1/sqrt3(hati+hatj+hatk)`
B. `1/sqrt5(sqrt2i-hatj-sqrt2k)`
C. `1/3(2hati+2hatj-hatk)`
D. `1/sqrt59(3hati-7hatj-hatk)`

Answer

To solve the problem, we need to find a unit vector \( \vec{a} \) such that the scalar triple product \( [\vec{a}, \vec{b}, \vec{c}] \) has the greatest value. The vectors 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: Find the cross product \( \vec{b} \times \vec{c} \)

To find the scalar triple product, we first need to calculate \( \vec{b} \times \vec{c} \).

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

Calculating the determinant:

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

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: Express the scalar triple product

The scalar triple product can be expressed as:

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

Let \( \vec{a} = x\hat{i} + y\hat{j} + z\hat{k} \), where \( x^2 + y^2 + z^2 = 1 \) (since \( \vec{a} \) is a unit vector).

Thus,

\[ [\vec{a}, \vec{b}, \vec{c}] = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 2\hat{j} - \hat{k}) = 2x + 2y - z \]

### Step 3: Maximize the expression \( 2x + 2y - z \)

To maximize \( 2x + 2y - z \) under the constraint \( x^2 + y^2 + z^2 = 1 \), we can use the method of Lagrange multipliers or substitute \( z = \sqrt{1 - x^2 - y^2} \).

Substituting \( z \):

\[ f(x, y) = 2x + 2y - \sqrt{1 - x^2 - y^2} \]

### Step 4: Find critical points

To find the maximum, we can analyze the function or check specific values for \( x, y, z \) that satisfy the unit vector condition.

### Step 5: Check possible values

We can check specific unit vectors. For example, if we try:

1. \( x = \frac{1}{\sqrt{3}}, y = \frac{1}{\sqrt{3}}, z = \frac{1}{\sqrt{3}} \)
2. \( x = \frac{\sqrt{2}}{\sqrt{5}}, y = -\frac{1}{\sqrt{5}}, z = -\frac{\sqrt{2}}{\sqrt{5}} \)
3. \( x = \frac{2}{\sqrt{3}}, y = \frac{2}{\sqrt{3}}, z = -\frac{1}{\sqrt{3}} \)
4. \( x = \frac{3}{\sqrt{59}}, y = -\frac{7}{\sqrt{59}}, z = -\frac{1}{\sqrt{59}} \)

After calculating the values, we find that the maximum occurs for the third option.

### Final Answer

Thus, the unit vector \( \vec{a} \) that maximizes the scalar triple product is:

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