Let `abs(a) = 3" and "abs(b)=4`. The value of `mu` for which the vectors `a+mub" and "a-mub` will be perpendicular is

Options

A. `3//4`
B. `2//3`
C. `-5//2`
D. `-2//3`

Answer

To find the value of \( \mu \) for which the vectors \( \mathbf{a} + \mu \mathbf{b} \) and \( \mathbf{a} - \mu \mathbf{b} \) are perpendicular, we can follow these steps:

### Step 1: Understand the condition for perpendicular vectors
Two vectors \( \mathbf{p} \) and \( \mathbf{q} \) are perpendicular if their dot product is zero:
\[ \mathbf{p} \cdot \mathbf{q} = 0 \]

### Step 2: Define the vectors
Let:
\[ \mathbf{p} = \mathbf{a} + \mu \mathbf{b} \]
\[ \mathbf{q} = \mathbf{a} - \mu \mathbf{b} \]

### Step 3: Set up the dot product
We need to find \( \mu \) such that:
\[ (\mathbf{a} + \mu \mathbf{b}) \cdot (\mathbf{a} - \mu \mathbf{b}) = 0 \]

### Step 4: Expand the dot product
Using the distributive property of the dot product:
\[ \mathbf{a} \cdot \mathbf{a} - \mu \mathbf{a} \cdot \mathbf{b} + \mu \mathbf{b} \cdot \mathbf{a} - \mu^2 \mathbf{b} \cdot \mathbf{b} = 0 \]
Since \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \), we can simplify this to:
\[ \mathbf{a} \cdot \mathbf{a} - \mu^2 \mathbf{b} \cdot \mathbf{b} = 0 \]

### Step 5: Substitute the magnitudes
Given \( |\mathbf{a}| = 3 \) and \( |\mathbf{b}| = 4 \):
\[ |\mathbf{a}|^2 = 3^2 = 9 \]
\[ |\mathbf{b}|^2 = 4^2 = 16 \]

### Step 6: Substitute these values into the equation
Now we have:
\[ 9 - \mu^2 \cdot 16 = 0 \]

### Step 7: Solve for \( \mu^2 \)
Rearranging gives:
\[ \mu^2 \cdot 16 = 9 \]
\[ \mu^2 = \frac{9}{16} \]

### Step 8: Find \( \mu \)
Taking the square root of both sides:
\[ \mu = \pm \frac{3}{4} \]

### Final Answer
The values of \( \mu \) for which the vectors are perpendicular are:
\[ \mu = \frac{3}{4} \quad \text{or} \quad \mu = -\frac{3}{4} \]