`[abi]i+[abj]j+[abk]k` is a equal to

Options

A. `axxb`
B. `a+b`
C. `a-b`
D. `bxxa`

Answer

To solve the problem `[abi]i + [abj]j + [abk]k`, we will break it down step by step.

### Step 1: Understand the notation
The notation `[abi]` represents the scalar triple product of vectors **a** and **b** with the unit vector **i**. Similarly, `[abj]` and `[abk]` represent the scalar triple products with unit vectors **j** and **k**, respectively.

### Step 2: Define the vectors
Let:
- **a** = \( a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \)
- **b** = \( b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \)

### Step 3: Calculate the scalar triple products
The scalar triple product can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors **a** and **b**.

1. **For `[abi]`:**
\[ [abi] = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1) \]

2. **For `[abj]`:**
\[ [abj] = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1) \]
The only difference is that we will take the dot product with **j**, which results in:
\[ [abj] = -(a_1b_3 - a_3b_1) \hat{j} \]

3. **For `[abk]`:**
\[ [abk] = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1) \]
The dot product with **k** results in:
\[ [abk] = (a_1b_2 - a_2b_1) \hat{k} \]

### Step 4: Combine the results
Now, we can combine the results of the three scalar triple products:
\[ [abi]i + [abj]j + [abk]k = (a_2b_3 - a_3b_2) \hat{i} - (a_1b_3 - a_3b_1) \hat{j} + (a_1b_2 - a_2b_1) \hat{k} \]

### Step 5: Final result
The final expression can be written as:
\[ = a \times b \]
where \( a \times b \) is the cross product of vectors **a** and **b**.

### Conclusion
Thus, the expression `[abi]i + [abj]j + [abk]k` is equal to the cross product \( a \times b \).

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