A vector perpendicular to `hat(i)+hat(j)+hat(k)` is

Options

A. `hat(i)-hat(j)+hat(k)`
B. `hat(i)-hat(j)-hat(k)`
C. `-hat(i)-hat(j)-hat(k)`
D. `3hat(i)+2hat(j)-5hat(k)`

Explanation

To find a vector that is perpendicular to the vector \(\hat{i} + \hat{j} + \hat{k}\), we can follow these steps:

### Step 1: Understand the Condition for Perpendicularity
Two vectors are perpendicular if their dot product is zero. If we denote the given vector as \(\mathbf{A} = \hat{i} + \hat{j} + \hat{k}\), we need to find another vector \(\mathbf{B}\) such that:
\[ \mathbf{A} \cdot \mathbf{B} = 0 \]

### Step 2: Define the Vector \(\mathbf{B}\)
Let's consider a general vector \(\mathbf{B} = a\hat{i} + b\hat{j} + c\hat{k}\), where \(a\), \(b\), and \(c\) are constants that we will determine.

### Step 3: Calculate the Dot Product
Now, we calculate the dot product \(\mathbf{A} \cdot \mathbf{B}\):
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (a\hat{i} + b\hat{j} + c\hat{k}) = a + b + c \]
We want this dot product to equal zero:
\[ a + b + c = 0 \]

### Step 4: Choose Values for \(a\), \(b\), and \(c\)
We can choose any values for \(a\), \(b\), and \(c\) that satisfy the equation \(a + b + c = 0\). For example, we can choose:
- \(a = 1\)
- \(b = -1\)
- \(c = 0\)

This gives us:
\[ \mathbf{B} = 1\hat{i} - 1\hat{j} + 0\hat{k} = \hat{i} - \hat{j} \]

### Step 5: Verify the Perpendicularity
Now, we verify that \(\mathbf{B} = \hat{i} - \hat{j}\) is indeed perpendicular to \(\mathbf{A}\):
\[ \mathbf{A} \cdot \mathbf{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j}) = 1 - 1 + 0 = 0 \]
Since the dot product is zero, \(\mathbf{B}\) is perpendicular to \(\mathbf{A}\).

### Conclusion
Thus, one vector that is perpendicular to \(\hat{i} + \hat{j} + \hat{k}\) is:
\[ \mathbf{B} = \hat{i} - \hat{j} \]

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