`a =i-2j+3k,b=3i+j+2k` then a vector c which is linear combination of a and b and also perpendicular to b is ...........

Answer

To find the vector \( \mathbf{c} \) which is a linear combination of vectors \( \mathbf{a} \) and \( \mathbf{b} \) and is also perpendicular to \( \mathbf{b} \), we can follow these steps:

### Step 1: Define the vectors
Given:
\[ \mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \]
\[ \mathbf{b} = 3\mathbf{i} + \mathbf{j} + 2\mathbf{k} \]

### Step 2: Express \( \mathbf{c} \) as a linear combination
Since \( \mathbf{c} \) is a linear combination of \( \mathbf{a} \) and \( \mathbf{b} \), we can express it as:
\[ \mathbf{c} = \mathbf{a} + \lambda \mathbf{b} \]
where \( \lambda \) is a scalar.

### Step 3: Substitute the vectors into the equation
Substituting the expressions for \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{c} = (\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) + \lambda(3\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \]
This simplifies to:
\[ \mathbf{c} = (1 + 3\lambda)\mathbf{i} + (-2 + \lambda)\mathbf{j} + (3 + 2\lambda)\mathbf{k} \]

### Step 4: Set up the condition for perpendicularity
For \( \mathbf{c} \) to be perpendicular to \( \mathbf{b} \), their dot product must be zero:
\[ \mathbf{c} \cdot \mathbf{b} = 0 \]
Calculating the dot product:
\[ (1 + 3\lambda)(3) + (-2 + \lambda)(1) + (3 + 2\lambda)(2) = 0 \]

### Step 5: Expand and simplify the equation
Expanding the dot product:
\[ 3(1 + 3\lambda) - 2 + \lambda + 2(3 + 2\lambda) = 0 \]
This simplifies to:
\[ 3 + 9\lambda - 2 + \lambda + 6 + 4\lambda = 0 \]
Combining like terms:
\[ (9\lambda + \lambda + 4\lambda) + (3 - 2 + 6) = 0 \]
\[ 14\lambda + 7 = 0 \]

### Step 6: Solve for \( \lambda \)
Solving for \( \lambda \):
\[ 14\lambda = -7 \implies \lambda = -\frac{1}{2} \]

### Step 7: Substitute \( \lambda \) back into \( \mathbf{c} \)
Now substitute \( \lambda = -\frac{1}{2} \) back into the expression for \( \mathbf{c} \):
\[ \mathbf{c} = (1 + 3(-\frac{1}{2}))\mathbf{i} + (-2 - \frac{1}{2})\mathbf{j} + (3 + 2(-\frac{1}{2}))\mathbf{k} \]
Calculating each component:
\[ \mathbf{c} = (1 - \frac{3}{2})\mathbf{i} + (-2 - \frac{1}{2})\mathbf{j} + (3 - 1)\mathbf{k} \]
\[ \mathbf{c} = -\frac{1}{2}\mathbf{i} - \frac{5}{2}\mathbf{j} + 2\mathbf{k} \]

### Step 8: Final result
Thus, the vector \( \mathbf{c} \) is:
\[ \mathbf{c} = -\frac{1}{2}\mathbf{i} - \frac{5}{2}\mathbf{j} + 2\mathbf{k} \]