If the vectors `2 hati +3 hat j +chatj` and `-3hat i+6hatk` are orthogonal, the value of c is :

Options

A. 0
B. `-1`
C. 1
D. `1//2`

Explanation

To find the value of \( c \) such that the vectors \( \mathbf{A} = 2\hat{i} + 3\hat{j} + c\hat{j} \) and \( \mathbf{B} = -3\hat{i} + 6\hat{k} \) are orthogonal, we can follow these steps:

### Step 1: Understand the condition for orthogonality
Two vectors are orthogonal if their dot product is zero. Therefore, we need to calculate the dot product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) and set it equal to zero.

### Step 2: Write down the vectors
The vectors are:
- \( \mathbf{A} = 2\hat{i} + (3 + c)\hat{j} \)
- \( \mathbf{B} = -3\hat{i} + 0\hat{j} + 6\hat{k} \)

### Step 3: Calculate the dot product
The dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (2\hat{i} + (3 + c)\hat{j}) \cdot (-3\hat{i} + 0\hat{j} + 6\hat{k}) \]
Using the properties of dot products:
\[ \mathbf{A} \cdot \mathbf{B} = 2 \cdot (-3) + (3 + c) \cdot 0 + 0 \cdot 6 \]
This simplifies to:
\[ \mathbf{A} \cdot \mathbf{B} = -6 + 0 + 0 = -6 \]

### Step 4: Include the \( \hat{k} \) component
Now, we need to include the \( \hat{k} \) component from vector \( \mathbf{A} \):
\[ \mathbf{A} = 2\hat{i} + (3 + c)\hat{j} + c\hat{k} \]
The dot product with \( \mathbf{B} \) now includes the \( \hat{k} \) component:
\[ \mathbf{A} \cdot \mathbf{B} = -6 + 6c \]

### Step 5: Set the dot product to zero
For the vectors to be orthogonal:
\[ -6 + 6c = 0 \]

### Step 6: Solve for \( c \)
Rearranging the equation gives:
\[ 6c = 6 \]
\[ c = 1 \]

### Conclusion
The value of \( c \) is \( 1 \).