For $\lambda>0$, 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

<p>We are given the vectors</p> <p>$ \vec{a} = \hat{i} + \lambda\,\hat{j} - 3\,\hat{k} \quad \text{and} \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}, $</p> <p>with $\lambda > 0$, and we are told that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular. Our goal is to find $(14 \cos\theta)^2$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.</p> <h3>Step 1. Use the Perpendicularity Condition</h3> <p>Two vectors are perpendicular if their dot product is zero. So, we require:</p> <p>$ (\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0. $</p> <p>First, compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$:</p> <p>$ \vec{a}+\vec{b} = (\hat{i}+3\,\hat{i})+(\lambda\,\hat{j}-\hat{j})+(-3\,\hat{k}+2\,\hat{k}) = 4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}. $</p> <p>$ \vec{a}-\vec{b} = (\hat{i}-3\,\hat{i})+(\lambda\,\hat{j}+ \hat{j})+(-3\,\hat{k}-2\,\hat{k}) = -2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}. $</p> <p>Now, take their dot product:</p> <p>$ (4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}) \cdot (-2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}) $</p> <p>$ = 4(-2) + (\lambda-1)(\lambda+1) + (-1)(-5) = -8 + (\lambda^2 - 1) + 5. $</p> <p>$ = \lambda^2 - 4. $</p> <p>Setting the dot product to zero:</p> <p>$ \lambda^2 - 4 = 0 \quad \Longrightarrow \quad \lambda^2 = 4. $</p> <p>Since $\lambda > 0$, we have:</p> <p>$ \lambda = 2. $</p> <h3>Step 2. Find the Angle Between $\vec{a}$ and $\vec{b}$</h3> <p>Now that we know $\lambda = 2$, the vectors become:</p> <p>$ \vec{a} = \hat{i} + 2\,\hat{j} - 3\,\hat{k}, \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}. $</p> <p>The cosine of the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by:</p> <p>$ \cos\theta = \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\| \|\vec{b}\|}. $</p> <p><strong>Calculate $\vec{a} \cdot \vec{b}$:</strong></p> <p>$ \vec{a}\cdot\vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5. $</p> <p><strong>Calculate the magnitudes:</strong></p> <p>$ \|\vec{a}\| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}. $</p> <p>$ \|\vec{b}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}. $</p> <p>Thus,</p> <p>$ \cos\theta = \frac{-5}{\sqrt{14}\cdot\sqrt{14}} = \frac{-5}{14}. $</p> <h3>Step 3. Compute $(14\cos\theta)^2$</h3> <p>$ (14\cos\theta)^2 = \left(14\cdot\frac{-5}{14}\right)^2 = (-5)^2 = 25. $</p> <h3>Final Answer</h3> <p>The value of $(14\cos\theta)^2$ is:</p> <p>$ \boxed{25}. $</p>