For `lambda gt 0`, let `theta` be the angle between the vectors `vec a = hat i + lambda hat j – 3 hat k` and `vec b = 3 hat i – hat j + 2 hat k`. If the vectors `vec a + vec b` and `vec a – vec b` are mutually perpendicular, then the value of `(14 cos theta)^2` is equal to

Options

A. 20
B. 25
C. 40
D. 50

Answer

To solve the problem step by step, we will follow the given conditions and perform the necessary calculations.

### Step 1: Define the vectors
Let
\[ \vec{a} = \hat{i} + \lambda \hat{j} - 3 \hat{k} \]
and
\[ \vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}. \]

### Step 2: Condition of perpendicularity
The vectors \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) are mutually perpendicular. Therefore, we can write:
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0. \]

### Step 3: Calculate \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\)
Calculating \(\vec{a} + \vec{b}\):
\[ \vec{a} + \vec{b} = (\hat{i} + \lambda \hat{j} - 3 \hat{k}) + (3 \hat{i} - \hat{j} + 2 \hat{k}) = (1 + 3) \hat{i} + (\lambda - 1) \hat{j} + (-3 + 2) \hat{k} = 4 \hat{i} + (\lambda - 1) \hat{j} - \hat{k}. \]

Calculating \(\vec{a} - \vec{b}\):
\[ \vec{a} - \vec{b} = (\hat{i} + \lambda \hat{j} - 3 \hat{k}) - (3 \hat{i} - \hat{j} + 2 \hat{k}) = (1 - 3) \hat{i} + (\lambda + 1) \hat{j} + (-3 - 2) \hat{k} = -2 \hat{i} + (\lambda + 1) \hat{j} - 5 \hat{k}. \]

### Step 4: Dot product of \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\)
Now, we find the dot product:
\[ (4 \hat{i} + (\lambda - 1) \hat{j} - \hat{k}) \cdot (-2 \hat{i} + (\lambda + 1) \hat{j} - 5 \hat{k}) = 0. \]

Calculating the dot product:
\[ 4 \cdot (-2) + (\lambda - 1)(\lambda + 1) + (-1)(-5) = 0. \]
This simplifies to:
\[ -8 + (\lambda^2 - 1) + 5 = 0. \]
Thus,
\[ \lambda^2 - 4 = 0. \]

### Step 5: Solve for \(\lambda\)
This gives:
\[ \lambda^2 = 4 \implies \lambda = 2 \quad (\text{since } \lambda > 0). \]

### Step 6: Substitute \(\lambda\) back into \(\vec{a}\) and \(\vec{b}\)
Now substituting \(\lambda = 2\):
\[ \vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}, \quad \vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}. \]

### Step 7: Calculate \(\cos \theta\)
To find \(\cos \theta\), we use the formula:
\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}. \]

Calculating \(\vec{a} \cdot \vec{b}\):
\[ \vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5. \]

Calculating \(|\vec{a}|\):
\[ |\vec{a}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}. \]

Calculating \(|\vec{b}|\):
\[ |\vec{b}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}. \]

### Step 8: Substitute into \(\cos \theta\)
Now substituting into the cosine formula:
\[ \cos \theta = \frac{-5}{\sqrt{14} \cdot \sqrt{14}} = \frac{-5}{14}. \]

### Step 9: Calculate \((14 \cos \theta)^2\)
Now we calculate:
\[ (14 \cos \theta)^2 = (14 \cdot \frac{-5}{14})^2 = (-5)^2 = 25. \]

### Final Answer
Thus, the value of \((14 \cos \theta)^2\) is:
\[ \boxed{25}. \]