Key Concepts and Formulas

1. Dot Product for Perpendicular Vectors: Two vectors $\vec{u}$ and $\vec{v}$ are mutually perpendicular if their dot product is zero: $\vec{u} \cdot \vec{v} = 0$.
2. Dot Product Identity: For any two vectors $\vec{a}$ and $\vec{b}$, the dot product of their sum and difference is $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$.
3. Magnitude of a Vector: For a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, its squared magnitude is $|\vec{v}|^2 = x^2 + y^2 + z^2$.
4. Angle Between Two Vectors: The cosine of the angle $\theta$ between two non-zero vectors $\vec{a}$ and $\vec{b}$ is given by $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$.

Step-by-Step Solution

Step 1: Determine the value of $\lambda$ using the perpendicularity condition. We are given that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are mutually perpendicular. This means their dot product is zero. $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = 0$ Using the dot product identity (Key Concept 2), this simplifies to: $|\vec{a}|^2 - |\vec{b}|^2 = 0$ $|\vec{a}|^2 = |\vec{b}|^2$ This implies that the magnitudes of vectors $\vec{a}$ and $\vec{b}$ are equal. Now, we calculate the squared magnitudes of $\vec{a}$ and $\vec{b}$ using Key Concept 3. For $\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$: $|\vec{a}|^2 = (1)^2 + (\lambda)^2 + (-3)^2 = 1 + \lambda^2 + 9 = \lambda^2 + 10$ For $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$: $|\vec{b}|^2 = (3)^2 + (-1)^2 + (2)^2 = 9 + 1 + 4 = 14$ Equating the squared magnitudes: $\lambda^2 + 10 = 14$ $\lambda^2 = 4$ Solving for $\lambda$, we get $\lambda = \pm 2$. Since the problem states that $\lambda > 0$, we take $\lambda = 2$.

Step 2: Calculate the dot product and magnitudes of $\vec{a}$ and $\vec{b}$ to find $\cos\theta$. With $\lambda=2$, the vector $\vec{a}$ is $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$. The vector $\vec{b}$ is $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$. We need to find the angle $\theta$ between $\vec{a}$ and $\vec{b}$. Using Key Concept 4, $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$. First, calculate the dot product $\vec{a}\cdot\vec{b}$: $\vec{a}\cdot\vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5$ Next, we find the magnitudes. From Step 1, we know $|\vec{a}|^2 = \lambda^2 + 10 = 2^2 + 10 = 4 + 10 = 14$, so $|\vec{a}| = \sqrt{14}$. Also from Step 1, we know $|\vec{b}|^2 = 14$, so $|\vec{b}| = \sqrt{14}$. Now, substitute these values into the formula for $\cos\theta$: $\cos\theta = \frac{-5}{\sqrt{14} \cdot \sqrt{14}} = \frac{-5}{14}$

Step 3: Compute the value of $(14 \cos \theta)^2$. We have found that $\cos\theta = -\frac{5}{14}$. We need to calculate $(14 \cos \theta)^2$. $(14 \cos \theta)^2 = \left(14 \cdot \left(-\frac{5}{14}\right)\right)^2$ The $14$ in the numerator and denominator cancel out: $(-5)^2 = 25$

Common Mistakes & Tips

• The identity $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$ is a crucial shortcut. Calculating $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ first and then their dot product would be more tedious.
• Pay close attention to the condition $\lambda > 0$ when solving for $\lambda$.
• Ensure accurate calculation of dot products and vector magnitudes, especially with signs.

Summary

The problem required us to first find the unknown scalar $\lambda$ by using the condition that $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular. This condition simplified to $|\vec{a}|^2 = |\vec{b}|^2$, which allowed us to solve for $\lambda$. Once $\lambda$ was determined, we calculated the dot product and magnitudes of $\vec{a}$ and $\vec{b}$ to find $\cos\theta$. Finally, we squared the expression $14 \cos \theta$ to obtain the answer.

The final answer is $\boxed{25}$.