Let `a=2i+2j+k` and b be another vector such that `a*b=14" and "a times b=3i+j-8k` then the vector b is equal to

Options

A. `5i+j+2k`
B. `5i-j-2k`
C. `5i+j-2k`
D. `3i+j+4k`

Answer

To find the vector \( \mathbf{b} \) given the conditions \( \mathbf{a} \cdot \mathbf{b} = 14 \) and \( \mathbf{a} \times \mathbf{b} = 3\mathbf{i} + \mathbf{j} - 8\mathbf{k} \), where \( \mathbf{a} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} \), we will follow these steps:

### Step 1: Define the vector \( \mathbf{b} \)
Let \( \mathbf{b} = \alpha \mathbf{i} + \beta \mathbf{j} + \gamma \mathbf{k} \).

### Step 2: Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \)
Using the formula for the dot product:
\[ \mathbf{a} \cdot \mathbf{b} = (2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \cdot (\alpha \mathbf{i} + \beta \mathbf{j} + \gamma \mathbf{k}) = 2\alpha + 2\beta + \gamma \]
We know from the problem statement that this equals 14:
\[ 2\alpha + 2\beta + \gamma = 14 \quad \text{(Equation 1)} \]

### Step 3: Calculate the cross product \( \mathbf{a} \times \mathbf{b} \)
Using the determinant method for the cross product:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ \alpha & \beta & \gamma \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{a} \times \mathbf{b} = \mathbf{i}(2\gamma - \beta) - \mathbf{j}(2\gamma - \alpha) + \mathbf{k}(2\beta - 2\alpha) \]
Setting this equal to \( 3\mathbf{i} + \mathbf{j} - 8\mathbf{k} \):
\[ 2\gamma - \beta = 3 \quad \text{(Equation 2)} \]
\[ -(2\gamma - \alpha) = 1 \implies 2\gamma - \alpha = -1 \quad \text{(Equation 3)} \]
\[ 2\beta - 2\alpha = -8 \implies \beta - \alpha = -4 \quad \text{(Equation 4)} \]

### Step 4: Solve the equations
From Equation 4, we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = \alpha - 4 \]

Substituting \( \beta \) into Equation 1:
\[ 2\alpha + 2(\alpha - 4) + \gamma = 14 \]
\[ 2\alpha + 2\alpha - 8 + \gamma = 14 \]
\[ 4\alpha + \gamma - 8 = 14 \]
\[ 4\alpha + \gamma = 22 \quad \text{(Equation 5)} \]

Now, substituting \( \beta \) into Equation 2:
\[ 2\gamma - (\alpha - 4) = 3 \]
\[ 2\gamma - \alpha + 4 = 3 \]
\[ 2\gamma - \alpha = -1 \quad \text{(Equation 6)} \]

### Step 5: Solve Equations 5 and 6
From Equation 6, we can express \( \gamma \) in terms of \( \alpha \):
\[ 2\gamma = \alpha - 1 \implies \gamma = \frac{\alpha - 1}{2} \]

Substituting \( \gamma \) into Equation 5:
\[ 4\alpha + \frac{\alpha - 1}{2} = 22 \]
Multiplying through by 2 to eliminate the fraction:
\[ 8\alpha + \alpha - 1 = 44 \]
\[ 9\alpha - 1 = 44 \]
\[ 9\alpha = 45 \implies \alpha = 5 \]

### Step 6: Find \( \beta \) and \( \gamma \)
Using \( \alpha = 5 \):
\[ \beta = 5 - 4 = 1 \]
\[ \gamma = \frac{5 - 1}{2} = 2 \]

### Step 7: Write the vector \( \mathbf{b} \)
Thus, the vector \( \mathbf{b} \) is:
\[ \mathbf{b} = 5\mathbf{i} + 1\mathbf{j} + 2\mathbf{k} \]

### Final Answer
The vector \( \mathbf{b} \) is \( 5\mathbf{i} + \mathbf{j} + 2\mathbf{k} \).

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