If `ahati +bhatj` is a unit vector and it is perpendicular to `hati +hatj` , then value of a and b is

Options

A. 1,0
B. `-2,0`
C. `0.5,-0.5`
D. None of these

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) such that the vector \( \vec{A} = a \hat{i} + b \hat{j} \) is a unit vector and is perpendicular to the vector \( \vec{B} = \hat{i} + \hat{j} \).

### Step-by-Step Solution:

1. **Understanding the Condition for Perpendicular Vectors**:
Two vectors are perpendicular if their dot product is zero. Therefore, we need to calculate the dot product of \( \vec{A} \) and \( \vec{B} \):
\[ \vec{A} \cdot \vec{B} = (a \hat{i} + b \hat{j}) \cdot (\hat{i} + \hat{j}) = 0 \]

2. **Calculating the Dot Product**:
Using the properties of dot products:
\[ \vec{A} \cdot \vec{B} = a (\hat{i} \cdot \hat{i}) + a (\hat{i} \cdot \hat{j}) + b (\hat{j} \cdot \hat{i}) + b (\hat{j} \cdot \hat{j}) \]
Since \( \hat{i} \cdot \hat{i} = 1 \), \( \hat{j} \cdot \hat{j} = 1 \), and \( \hat{i} \cdot \hat{j} = 0 \):
\[ \vec{A} \cdot \vec{B} = a \cdot 1 + b \cdot 1 = a + b \]
Setting this equal to zero gives us:
\[ a + b = 0 \quad \text{(1)} \]

3. **Condition for Unit Vector**:
A vector is a unit vector if its magnitude is 1. The magnitude of \( \vec{A} \) is given by:
\[ |\vec{A}| = \sqrt{a^2 + b^2} = 1 \]
Squaring both sides, we get:
\[ a^2 + b^2 = 1 \quad \text{(2)} \]

4. **Substituting from Equation (1)**:
From equation (1), we can express \( b \) in terms of \( a \):
\[ b = -a \]
Substituting this into equation (2):
\[ a^2 + (-a)^2 = 1 \]
Simplifying, we have:
\[ a^2 + a^2 = 1 \implies 2a^2 = 1 \implies a^2 = \frac{1}{2} \]
Taking the square root gives:
\[ a = \pm \frac{1}{\sqrt{2}} \quad \text{and thus} \quad b = \mp \frac{1}{\sqrt{2}} \]

5. **Final Values**:
Therefore, the values of \( a \) and \( b \) can be:
\[ (a, b) = \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \quad \text{or} \quad (a, b) = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \]