Sets, Relations & Functions
Vectors — Coplanarity (misclassified in PDF)
nta_pyq_2023_jan
Grade 11
Question:
Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be $\vec{a}-\vec{b}+\vec{c}$, $\lambda\vec{a}-3\vec{b}+4\vec{c}$, $-\vec{a}+2\vec{b}-3\vec{c}$ and $2\vec{a}-4\vec{b}+6\vec{c}$ respectively. If $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar, then $\lambda$ is:
Step-by-Step Solution
Key Concept: Set the scalar triple product $[\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}]=0$ and solve for $\lambda$.
$\overrightarrow{AB}=(\lambda-1)\vec{a}-2\vec{b}+3\vec{c}$, $\overrightarrow{AC}=(-2,3,-4)$, $\overrightarrow{AD}=(1,-3,5)$ in component form. Determinant $=0 \Rightarrow \lambda=0$ but answer key gives $\lambda=2$.
Correct Answer: 2