The dot product of a vector with the vectors `hati + hatj - 3 hatk , hati + 3 hatj - 2 hatk and 2 hati + hatj + 4 hatk are 0,5 and 8` respectively. The vector is

Options

A. `hati + 2 hatj + hatk `
B. `- hati + 3 hatj - 2 hatk `
C. `hati + 2 hatj + 3 hatk `
D. `hati - 3 hatj - 3 hatk`

Answer

To find the vector \( \mathbf{a} \) that satisfies the given dot products with the specified vectors, we can follow these steps:

### Step 1: Define the Vector
Let the vector \( \mathbf{a} \) be represented as:
\[ \mathbf{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \]

### Step 2: Set Up the Dot Product Equations
We are given three dot product equations based on the problem statement:
1. \( \mathbf{a} \cdot (\hat{i} + \hat{j} - 3\hat{k}) = 0 \)
2. \( \mathbf{a} \cdot (\hat{i} + 3\hat{j} - 2\hat{k}) = 5 \)
3. \( \mathbf{a} \cdot (2\hat{i} + \hat{j} + 4\hat{k}) = 8 \)

### Step 3: Write the Equations
From the first equation:
\[ a_1 + a_2 - 3a_3 = 0 \quad \text{(Equation 1)} \]

From the second equation:
\[ a_1 + 3a_2 - 2a_3 = 5 \quad \text{(Equation 2)} \]

From the third equation:
\[ 2a_1 + a_2 + 4a_3 = 8 \quad \text{(Equation 3)} \]

### Step 4: Solve the Equations
We will solve these equations step by step.

#### Subtract Equation 1 from Equation 2:
\[ (a_1 + 3a_2 - 2a_3) - (a_1 + a_2 - 3a_3) = 5 - 0 \]
This simplifies to:
\[ 2a_2 + a_3 = 5 \quad \text{(Equation 4)} \]

#### Subtract Equation 1 from Equation 3:
\[ (2a_1 + a_2 + 4a_3) - (a_1 + a_2 - 3a_3) = 8 - 0 \]
This simplifies to:
\[ a_1 + 7a_3 = 8 \quad \text{(Equation 5)} \]

### Step 5: Express Variables
From Equation 4, we can express \( a_3 \) in terms of \( a_2 \):
\[ a_3 = 5 - 2a_2 \quad \text{(Substituting into Equation 5)} \]

Substituting \( a_3 \) into Equation 5:
\[ a_1 + 7(5 - 2a_2) = 8 \]
This simplifies to:
\[ a_1 + 35 - 14a_2 = 8 \]
Thus,
\[ a_1 = 14a_2 - 27 \quad \text{(Equation 6)} \]

### Step 6: Substitute Back
Now substitute Equation 6 into Equation 1:
\[ (14a_2 - 27) + a_2 - 3(5 - 2a_2) = 0 \]
Expanding this gives:
\[ 14a_2 - 27 + a_2 - 15 + 6a_2 = 0 \]
Combining like terms:
\[ 21a_2 - 42 = 0 \]
Thus,
\[ a_2 = 2 \]

### Step 7: Find \( a_3 \) and \( a_1 \)
Substituting \( a_2 = 2 \) back into Equation 4:
\[ 2a_3 = 5 - 2(2) \Rightarrow 2a_3 = 1 \Rightarrow a_3 = \frac{1}{2} \]

Now substituting \( a_2 = 2 \) into Equation 6:
\[ a_1 = 14(2) - 27 = 28 - 27 = 1 \]

### Final Vector
Thus, the vector \( \mathbf{a} \) is:
\[ \mathbf{a} = 1\hat{i} + 2\hat{j} + \frac{1}{2}\hat{k} \]

### Conclusion
The vector \( \mathbf{a} \) that satisfies the given conditions is:
\[ \mathbf{a} = \hat{i} + 2\hat{j} + \frac{1}{2}\hat{k} \]