Vector Algebra
Cross Product
Grade 12
Question:
<p>Let <span>\(\vec{a} = \vec{i} + 2\vec{j} - 3\vec{k}\)</span> and <span>\(\vec{b} = 2\vec{i} - 3\vec{j} + 5\vec{k}\)</span>. If <span>\(\vec{r} \times \vec{a} = \vec{b} \times \vec{r}\)</span>, <span>\(\vec{r} \times (a\vec{i} + 2\vec{j} + \vec{k}) = 3\)</span> and <span>\(\vec{r} \times (2\vec{i} + 5\vec{j} - a\vec{k}) = -1\)</span>, where <span>\(a \in \mathbb{R}\)</span>, then the value of <span>\(a + |\vec{r}|^2\)</span> is equal to</p>
<p>(a) 9</p>
<p>(b) 15</p>
<p>(c) 13</p>
<p>(d) 11</p>
Step-by-Step Solution
Key Concept: When the cross product condition is satisfied, the position vector r must be parallel to the sum of the given vectors, which constrains the solution.
Step 1: From $\vec{r} \times \vec{a} = \vec{b} \times \vec{r}$ , we get $\vec{r} \times (\vec{a} + \vec{b}) = \vec{0}$ Step 2: This implies $\vec{r} = \lambda(\vec{a} + \vec{b})$ for some scalar $\lambda$ $\vec{r} = \lambda(\hat{i} + 2\hat{j} - 3\hat{k} + 2\hat{i} - 3\hat{j} + 5\hat{k})$ $\vec{r} = \lambda(3\hat{i} - \hat{j} + 2\hat{k})$ Step 3: Using the condition $\vec{r} \times (a\hat{i} + 2\hat{j} + \hat{k}) = 3$ and substituting the expression for $\vec{r}$ , we can determine $\lambda$ and $a$ . ∴ Answer is (b) 15.
Correct Answer: B