Consider a vector `vecF=4hati-3hatj`. Another vector that is perpendicular to `vecF` is

Options

A. `4hati+3hatj`
B. `6hati`
C. `7hatk`
D. `3hati -4hatj`

Explanation

To find a vector that is perpendicular to the given vector \(\vec{F} = 4\hat{i} - 3\hat{j}\), we can use the property that two vectors are perpendicular if their dot product is zero.

### Step-by-Step Solution:

1. **Identify the Given Vector**:
The vector given is \(\vec{F} = 4\hat{i} - 3\hat{j}\).

2. **Set Up the Perpendicular Condition**:
Let the vector we are looking for be \(\vec{A} = a\hat{i} + b\hat{j}\). For \(\vec{A}\) to be perpendicular to \(\vec{F}\), the dot product must equal zero:
\[ \vec{F} \cdot \vec{A} = 0 \]

3. **Calculate the Dot Product**:
The dot product \(\vec{F} \cdot \vec{A}\) is calculated as follows:
\[ \vec{F} \cdot \vec{A} = (4\hat{i} - 3\hat{j}) \cdot (a\hat{i} + b\hat{j}) = 4a - 3b \]

4. **Set the Dot Product to Zero**:
To satisfy the perpendicular condition, we set the dot product to zero:
\[ 4a - 3b = 0 \]

5. **Solve for One Variable**:
Rearranging the equation gives:
\[ 4a = 3b \quad \Rightarrow \quad b = \frac{4}{3}a \]
This means that for any value of \(a\), we can find a corresponding \(b\) that will make \(\vec{A}\) perpendicular to \(\vec{F}\).

6. **Choose a Specific Value**:
For simplicity, let’s choose \(a = 3\). Then:
\[ b = \frac{4}{3} \times 3 = 4 \]
So one possible vector is:
\[ \vec{A} = 3\hat{i} + 4\hat{j} \]

7. **Check the Options**:
Now, let’s check the given options:
- Option 1: \(4\hat{i} + 3\hat{j}\)
- Option 2: \(6\hat{i}\)
- Option 3: \(7\hat{k}\)
- Option 4: \(3\hat{i} - 4\hat{j}\)

The vector \(7\hat{k}\) (Option 3) is also perpendicular to any vector in the \(i-j\) plane, including \(\vec{F}\).

### Conclusion:
Thus, both \(3\hat{i} + 4\hat{j}\) and \(7\hat{k}\) are vectors that are perpendicular to \(\vec{F}\). However, since the question asks for a specific vector from the options provided, the correct answer is **Option 3: \(7\hat{k}\)**.