Vector Algebra
Cross product and scalar product
Grade 12
Question:
<p>Given \(\vec{a} = \hat{i} - \hat{j}\), \(\vec{b} = \hat{i} + \hat{j} + \hat{k}\), \(\vec{a} \times \vec{c} + \vec{b} = \vec{0}\) and \(\vec{a} \cdot \vec{c} = 4\). Find \(|\vec{c}|^2\).</p>
<p>\(8\)</p>
<p>\(\dfrac{19}{2}\)</p>
<p>\(9\)</p>
<p>\(\dfrac{17}{2}\)</p>
Step-by-Step Solution
Key Concept: From the constraint $\vec{a} \times \vec{c} + \vec{b} = \vec{0}$, we get $\vec{a} \times \vec{c} = -\vec{b}$. Taking the magnitude squared and using the dot product constraint $\vec{a} \cdot \vec{c} = 4$, we can solve for $|\vec{c}|^2$.
Step 1: Express $\vec{c}$ using the constraint $\vec{a} \times \vec{c} = -\vec{b}$ From $\vec{a} \times \vec{c} + \vec{b} = \vec{0}$, we have $\vec{a} \times \vec{c} = -\vec{b}$ Taking magnitude squared on both sides: $|\vec{a} \times \vec{c}|^2 = |-\vec{b}|^2 = |\vec{b}|^2$ Step 2: Calculate $|\vec{a}|^2$ and $|\vec{b}|^2$ $\vec{a} = \hat{i} - \hat{j}$ gives $|\vec{a}|^2 = 1 + 1 = 2$ $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ gives $|\vec{b}|^2 = 1 + 1 + 1 = 3$ Step 3: Apply Lagrange's Identity Using $|\vec{a} \times \vec{c}|^2 = |\vec{a}|^2|\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2$: $3 = 2|\vec{c}|^2 - (4)^2$ $3 = 2|\vec{c}|^2 - 16$ Step 4: Solve for $|\vec{c}|^2$ $2|\vec{c}|^2 = 19$ $|\vec{c}|^2 = \frac{19}{2}$ ∴ Answer: B
Correct Answer: B