When vector `hat(n) = ahat(i) + bhat(j)` is perpendicular to `(2hat(i) + hat(j))`, then a and b are

Options

A. `(1)/(sqrt5),(-2)/(sqrt5)`
B. `-2,0`
C. 0,-2
D. `(1)/(sqrt2),-(1)/(sqrt2)`

Answer

To solve the problem where the vector \(\hat{n} = a\hat{i} + b\hat{j}\) is perpendicular to the vector \(2\hat{i} + \hat{j}\), we can follow these steps:

### Step 1: Understand the condition for perpendicularity
Two vectors are perpendicular if their dot product is zero. Therefore, we can write the equation:
\[ \hat{n} \cdot (2\hat{i} + \hat{j}) = 0 \]

### Step 2: Calculate the dot product
Substituting \(\hat{n} = a\hat{i} + b\hat{j}\) into the dot product, we get:
\[ (a\hat{i} + b\hat{j}) \cdot (2\hat{i} + \hat{j}) = 0 \]
Calculating the dot product:
\[ a \cdot 2 + b \cdot 1 = 2a + b = 0 \]

### Step 3: Set up the first equation
From the dot product, we have our first equation:
\[ 2a + b = 0 \quad \text{(Equation 1)} \]

### Step 4: Use the unit vector condition
Since \(\hat{n}\) is a unit vector, its magnitude must equal 1. Thus, we can write:
\[ \sqrt{a^2 + b^2} = 1 \]
Squaring both sides gives:
\[ a^2 + b^2 = 1 \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 1 into Equation 2
From Equation 1, we can express \(b\) in terms of \(a\):
\[ b = -2a \]
Now substitute this into Equation 2:
\[ a^2 + (-2a)^2 = 1 \]
This simplifies to:
\[ a^2 + 4a^2 = 1 \]
\[ 5a^2 = 1 \]

### Step 6: Solve for \(a\)
Dividing both sides by 5 gives:
\[ a^2 = \frac{1}{5} \]
Taking the square root:
\[ a = \pm \frac{1}{\sqrt{5}} \]

### Step 7: Find \(b\)
Using \(b = -2a\):
- If \(a = \frac{1}{\sqrt{5}}\), then:
\[ b = -2\left(\frac{1}{\sqrt{5}}\right) = -\frac{2}{\sqrt{5}} \]
- If \(a = -\frac{1}{\sqrt{5}}\), then:
\[ b = -2\left(-\frac{1}{\sqrt{5}}\right) = \frac{2}{\sqrt{5}} \]

### Final Result
Thus, the pairs \((a, b)\) can be:
1. \(a = \frac{1}{\sqrt{5}}, b = -\frac{2}{\sqrt{5}}\)
2. \(a = -\frac{1}{\sqrt{5}}, b = \frac{2}{\sqrt{5}}\)