Vector Algebra
Coplanar vectors and cross product
Grade 12
Question:
<p>Given \(\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}\), \(\vec{b} = \hat{i} + \lambda \hat{j} + 4\hat{k}\) and \(\vec{c} = 2\hat{i} + 4\hat{j} + (\lambda^2 - 1)\hat{k}\) are coplanar vectors. Find \(\vec{a} \times \vec{c}\).</p>
<p>\(-10\hat{i} + 5\hat{j}\)</p>
<p>\(10\hat{i} - 5\hat{j}\)</p>
<p>\(-10\hat{i} - 5\hat{j}\)</p>
<p>\(10\hat{i} + 5\hat{j}\)</p>
Step-by-Step Solution
Key Concept: Three vectors are coplanar if and only if their scalar triple product equals zero: $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$. This condition gives a determinant equation that determines $\lambda$, allowing you to find $\vec{c}$ and then compute $\vec{a} \times \vec{c}$.
Step 1: Apply coplanarity condition For coplanar vectors: $\begin{vmatrix} 1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & \lambda^2-1 \end{vmatrix} = 0$ Step 2: Expand the determinant $1(\lambda(\lambda^2-1) - 16) - 2(1(\lambda^2-1) - 8) + 4(4 - 2\lambda) = 0$ $\lambda^3 - \lambda - 16 - 2\lambda^2 + 2 + 16 + 16 - 8\lambda = 0$ $\lambda^3 - 2\lambda^2 - 9\lambda + 18 = 0$ Step 3: Solve for $\lambda$ Factoring: $(\lambda - 2)(\lambda^2 - 9) = 0$ $(\lambda - 2)(\lambda - 3)(\lambda + 3) = 0$ So $\lambda = 2, 3,$ or $-3$ Step 4: Calculate $\vec{a} \times \vec{c}$ for valid $\lambda$ For $\lambda = 2$: $\vec{c} = 2\hat{i} + 4\hat{j} + 3\hat{k}$ $\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \end{vmatrix} = \hat{i}(6-16) - \hat{j}(3-8) + \hat{k}(4-4)$ $= -10\hat{i} + 5\hat{j} + 0\hat{k} = -10\hat{i} + 5\hat{j}$ ∴ Answer: A ($-10\hat{i} + 5\hat{j}$ or equivalent form)
Correct Answer: A