Vector Algebra
Cross Product and Perpendicular Vectors
Grade 12
Question:
<p><strong>65.</strong> Let \(\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\) be two vectors. If a vector perpendicular to both the vectors \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) has the magnitude 12, then one such vector is:</p>
<p>\(4(2\hat{i} + 2\hat{j} + \hat{k})\)</p>
<p>\(4(2\hat{i} - 2\hat{j} - \hat{k})\)</p>
<p>\(4(2\hat{i} + 2\hat{j} - \hat{k})\)</p>
<p>\(4(-2\hat{i} - 2\hat{j} + \hat{k})\)</p>
Step-by-Step Solution
Key Concept: A vector perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is found using the cross product, then scaled to magnitude 12. Note that $(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = 2(\vec{a} \times \vec{b})$ by the vector identity.
Step 1: Compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$: $\vec{a}+\vec{b} = (3\hat{i}+2\hat{j}+2\hat{k}) + (\hat{i}+2\hat{j}-2\hat{k}) = 4\hat{i}+4\hat{j}+0\hat{k}$ $\vec{a}-\vec{b} = (3\hat{i}+2\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}-2\hat{k}) = 2\hat{i}+0\hat{j}+4\hat{k}$ Step 2: Find the cross product $(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})$: $(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = \hat{i}(16-0) - \hat{j}(16-0) + \hat{k}(0-8)$ $= 16\hat{i} - 16\hat{j} - 8\hat{k}$ Step 3: Find the magnitude of this cross product: $|(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256+256+64} = \sqrt{576} = 24$ Step 4: Scale to magnitude 12: Required vector $= \frac{12}{24}(16\hat{i} - 16\hat{j} - 8\hat{k}) = \frac{1}{2}(16\hat{i} - 16\hat{j} - 8\hat{k}) = 8\hat{i} - 8\hat{j} - 4\hat{k}$ ∴ Answer: A
Correct Answer: A