Let `vec a=-hat i-hat j+hat k,vec a*vec b=1` and `vec a timesvec b=hat i-hat j`.Then `vec a-6vec b` is equal to

Options

A. `3(hat i-hat j-hat k)`
B. `3(hat i-hat j+hat k)`
C. `3(hat i+hat j-hat k),`
D. `3(hat i+hat j+hat k)`

Answer

To solve the problem, we need to find the expression for \( \vec{a} - 6\vec{b} \) given the vectors and their relationships. Let's break down the solution step by step.

### Step 1: Define the vectors
We are given:
\[ \vec{a} = -\hat{i} - \hat{j} + \hat{k} \]
Let \( \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \).

### Step 2: Use the dot product condition
We know that:
\[ \vec{a} \cdot \vec{b} = 1 \]
Calculating the dot product:
\[ (-1)(x) + (-1)(y) + (1)(z) = 1 \]
This simplifies to:
\[ -z - x - y = 1 \quad \text{(Equation 1)} \]

### Step 3: Use the cross product condition
We also know that:
\[ \vec{a} \times \vec{b} = \hat{i} - \hat{j} \]
Calculating the cross product using the determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 1 \\ x & y & z \end{vmatrix} \]
Calculating this determinant:
\[ = \hat{i} \left((-1)z - (1)(y)\right) - \hat{j} \left((-1)z - (1)(x)\right) + \hat{k} \left((-1)(y) - (-1)(x)\right) \]
This simplifies to:
\[ = \hat{i}(-z - y) - \hat{j}(-z - x) + \hat{k}(-y + x) \]
Setting this equal to \( \hat{i} - \hat{j} \):
\[ -z - y = 1 \quad \text{(Equation 2)} \]
\[ -z - x = -1 \quad \text{(Equation 3)} \]
\[ -x + y = 0 \quad \text{(Equation 4)} \]

### Step 4: Solve the equations
From Equation 4, we have:
\[ y = x \]
Substituting \( y = x \) into Equation 1:
\[ -z - x - x = 1 \Rightarrow -z - 2x = 1 \Rightarrow z = -2x - 1 \quad \text{(Equation 5)} \]
Substituting \( y = x \) into Equation 2:
\[ -z - x = 1 \Rightarrow -z = 1 + x \Rightarrow z = -1 - x \quad \text{(Equation 6)} \]
Equating Equations 5 and 6:
\[ -2x - 1 = -1 - x \]
Solving this gives:
\[ -2x = -x \Rightarrow x = 0 \]
Then substituting \( x = 0 \) back into Equation 4 gives:
\[ y = 0 \]
And substituting \( x = 0 \) into Equation 6 gives:
\[ z = -1 \]
Thus, we have:
\[ \vec{b} = 0\hat{i} + 0\hat{j} - 1\hat{k} = -\hat{k} \]

### Step 5: Calculate \( \vec{a} - 6\vec{b} \)
Now substituting \( \vec{b} \) into the expression:
\[ \vec{a} - 6\vec{b} = (-\hat{i} - \hat{j} + \hat{k}) - 6(-\hat{k}) \]
This simplifies to:
\[ = -\hat{i} - \hat{j} + \hat{k} + 6\hat{k} = -\hat{i} - \hat{j} + 7\hat{k} \]

### Final Result
Thus, the final result is:
\[ \vec{a} - 6\vec{b} = -\hat{i} - \hat{j} + 7\hat{k} \]