Vector Algebra
Cross Product and Perpendicular Vectors
Grade 12
Question:
<p>Given \(\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\). A vector perpendicular to both \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) is</p>
<p>\(8\lambda(2\hat{i} - 2\hat{j} - \hat{k})\)</p>
<p>\(8\lambda(2\hat{i} - 2\hat{j} + \hat{k})\)</p>
<p>\(8\lambda(2\hat{i} + 2\hat{j} - \hat{k})\)</p>
<p>\(8\lambda(-2\hat{i} + 2\hat{j} - \hat{k})\)</p>
Step-by-Step Solution
Key Concept: A vector perpendicular to both given vectors is their cross product. First simplify by finding $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, then compute the cross product to get a perpendicular vector.
Step 1: Calculate $\vec{a} + \vec{b}$ 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: Simplify by factoring out 8 $= 8(2\hat{i} - 2\hat{j} - \hat{k})$ ∴ Answer: A (The vector $2\hat{i} - 2\hat{j} - \hat{k}$ or any scalar multiple is perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$)
Correct Answer: A