If `veca` satisfies `vecaxx(hati+2hatj+hatk)=hati-hatk" then " veca` is equal to

Options

A. a) `lambdahati+(2lambda-1)hatj+lambdahatk,lambda in R`
B. b) `lambdahati+(1-2lambda)hatj+lambdahatk,lambda in R`
C. c) `lambdahati+(2lambda+1)hatj+lambdahatk,lambda in R`
D. d) `lambdahati+(1+2lambda)hatj+lambdahatk,lambda in R`

Explanation

To solve the given problem, we need to find the vector \(\vec{a}\) that satisfies the equation:

\[ \vec{a} \times \vec{v} = \hat{i} - \hat{k} \]

where \(\vec{v} = \hat{i} + 2\hat{j} + \hat{k}\).

### Step 1: Write the cross product equation

We start with the equation:

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

### Step 2: Use the properties of cross products

Recall that the cross product of vectors has the following properties:

- \(\hat{i} \times \hat{i} = \vec{0}\)
- \(\hat{j} \times \hat{j} = \vec{0}\)
- \(\hat{k} \times \hat{k} = \vec{0}\)
- \(\hat{j} \times \hat{i} = -\hat{k}\)
- \(\hat{k} \times \hat{j} = \hat{i}\)
- \(\hat{i} \times \hat{k} = \hat{j}\)

### Step 3: Express \(\vec{a}\) in terms of its components

Let \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\). Then we can expand the left-hand side:

\[ \vec{a} \times (\hat{i} + 2\hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ 1 & 2 & 1 \end{vmatrix} \]

### Step 4: Calculate the determinant

Calculating the determinant, we have:

\[ \vec{a} \times (\hat{i} + 2\hat{j} + \hat{k}) = \hat{i}(a_2 \cdot 1 - a_3 \cdot 2) - \hat{j}(a_1 \cdot 1 - a_3 \cdot 1) + \hat{k}(a_1 \cdot 2 - a_2 \cdot 1) \]

This simplifies to:

\[ \vec{a} \times (\hat{i} + 2\hat{j} + \hat{k}) = (a_2 - 2a_3)\hat{i} - (a_1 - a_3)\hat{j} + (2a_1 - a_2)\hat{k} \]

### Step 5: Set the components equal to \(\hat{i} - \hat{k}\)

Now, we equate the components from both sides:

1. \(a_2 - 2a_3 = 1\) (coefficient of \(\hat{i}\))
2. \(-(a_1 - a_3) = 0\) (coefficient of \(\hat{j}\))
3. \(2a_1 - a_2 = -1\) (coefficient of \(\hat{k}\))

### Step 6: Solve the system of equations

From the second equation, we have:

\[ a_1 = a_3 \]

Substituting \(a_1 = a_3\) into the first equation:

\[ a_2 - 2a_1 = 1 \implies a_2 = 1 + 2a_1 \]

Now substituting \(a_1\) and \(a_2\) into the third equation:

\[ 2a_1 - (1 + 2a_1) = -1 \]

This simplifies to:

\[ 2a_1 - 1 - 2a_1 = -1 \implies -1 = -1 \]

This is always true, meaning \(a_1\) can take any value. Let \(a_1 = \lambda\), then:

- \(a_3 = \lambda\)
- \(a_2 = 1 + 2\lambda\)

### Step 7: Write the final expression for \(\vec{a}\)

Thus, we can write:

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

where \(\lambda\) is any real number.

### Final Answer:

\[ \vec{a} = \lambda \hat{i} + (1 + 2\lambda) \hat{j} + \lambda \hat{k}, \quad \lambda \in \mathbb{R} \]

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