If magnitude of sum of two unit vectors is `sqrt2` then find the magnitude of subtraction of these unit vectors.

Options

A. √
2
B. √
5
C. 2
D. √
7

Explanation

To solve the problem step by step, we will use the properties of vectors and the information given about their magnitudes.

### Step 1: Understand the given information
We are given that the magnitude of the sum of two unit vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \sqrt{2} \). Since both \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors, their magnitudes are \( |\mathbf{a}| = 1 \) and \( |\mathbf{b}| = 1 \).

### Step 2: Write the equation for the magnitude of the sum
The magnitude of the sum of two vectors can be expressed using the formula:
\[ |\mathbf{a} + \mathbf{b}| = \sqrt{|\mathbf{a}|^2 + |\mathbf{b}|^2 + 2 |\mathbf{a}||\mathbf{b}|\cos\theta} \]
where \( \theta \) is the angle between the two vectors.

### Step 3: Substitute the known values
Since \( |\mathbf{a}| = 1 \) and \( |\mathbf{b}| = 1 \), we can substitute these values into the equation:
\[ |\mathbf{a} + \mathbf{b}| = \sqrt{1^2 + 1^2 + 2 \cdot 1 \cdot 1 \cdot \cos\theta} \]
This simplifies to:
\[ |\mathbf{a} + \mathbf{b}| = \sqrt{2 + 2\cos\theta} \]

### Step 4: Set the equation equal to the given magnitude
We know that \( |\mathbf{a} + \mathbf{b}| = \sqrt{2} \). Therefore, we can set up the equation:
\[ \sqrt{2 + 2\cos\theta} = \sqrt{2} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 2 + 2\cos\theta = 2 \]

### Step 6: Solve for \( \cos\theta \)
Subtracting 2 from both sides results in:
\[ 2\cos\theta = 0 \]
This implies:
\[ \cos\theta = 0 \]
Thus, \( \theta = 90^\circ \) (the vectors are perpendicular).

### Step 7: Find the magnitude of the subtraction of the vectors
Now, we need to find the magnitude of the subtraction \( |\mathbf{a} - \mathbf{b}| \). We use the formula:
\[ |\mathbf{a} - \mathbf{b}| = \sqrt{|\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta} \]
Substituting the values:
\[ |\mathbf{a} - \mathbf{b}| = \sqrt{1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \cos(90^\circ)} \]
Since \( \cos(90^\circ) = 0 \), we have:
\[ |\mathbf{a} - \mathbf{b}| = \sqrt{1 + 1 - 0} = \sqrt{2} \]

### Final Answer
Thus, the magnitude of the subtraction of the two unit vectors is:
\[ |\mathbf{a} - \mathbf{b}| = \sqrt{2} \]