If `|vec(a)|=3, |vec(b)|+4`, then for what value of 1 is `(vec(a)+lambdavec(b))` perpendicular to `(vec(a)-lambdavec(b))`?

Options

A. `+-3/4`
B. `+-4/3`
C. `+-9/16`
D. `+-3/5`

Explanation

To solve the problem, we need to find the value of \( \lambda \) such that the vector \( \vec{a} + \lambda \vec{b} \) is perpendicular to the vector \( \vec{a} - \lambda \vec{b} \).

### Step-by-step Solution:

1. **Understanding Perpendicular Vectors**:
Two vectors \( \vec{u} \) and \( \vec{v} \) are perpendicular if their dot product is zero:
\[ \vec{u} \cdot \vec{v} = 0 \]
In our case, we have:
\[ (\vec{a} + \lambda \vec{b}) \cdot (\vec{a} - \lambda \vec{b}) = 0 \]

2. **Calculating the Dot Product**:
We can expand the dot product:
\[ (\vec{a} + \lambda \vec{b}) \cdot (\vec{a} - \lambda \vec{b}) = \vec{a} \cdot \vec{a} - \lambda \vec{a} \cdot \vec{b} + \lambda \vec{b} \cdot \vec{a} - \lambda^2 \vec{b} \cdot \vec{b} \]
Since \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \), this simplifies to:
\[ \vec{a} \cdot \vec{a} - \lambda^2 \vec{b} \cdot \vec{b} \]

3. **Substituting Magnitudes**:
We know the magnitudes:
\[ |\vec{a}| = 3 \quad \text{and} \quad |\vec{b}| = 4 \]
Therefore:
\[ \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 3^2 = 9 \]
\[ \vec{b} \cdot \vec{b} = |\vec{b}|^2 = 4^2 = 16 \]

4. **Setting Up the Equation**:
Now we can substitute these values into our dot product equation:
\[ 9 - \lambda^2 \cdot 16 = 0 \]

5. **Solving for \( \lambda^2 \)**:
Rearranging gives:
\[ 9 = 16\lambda^2 \]
Dividing both sides by 16:
\[ \lambda^2 = \frac{9}{16} \]

6. **Finding \( \lambda \)**:
Taking the square root of both sides:
\[ \lambda = \pm \frac{3}{4} \]

### Final Answer:
Thus, the values of \( \lambda \) for which \( \vec{a} + \lambda \vec{b} \) is perpendicular to \( \vec{a} - \lambda \vec{b} \) are:
\[ \lambda = \frac{3}{4} \quad \text{or} \quad \lambda = -\frac{3}{4} \]