Vector Algebra
Coplanarity condition for vectors
nta_pyq_2023_jan
Grade 12
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: Compute $\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}$ in terms of $\vec{a},\vec{b},\vec{c}$, set their scalar triple product $=0$ (determinant of coefficients).
$\overrightarrow{AB} = (\lambda-1)\vec{a}-2\vec{b}+3\vec{c}$, $\overrightarrow{AC} = -2\vec{a}+3\vec{b}-4\vec{c}$, $\overrightarrow{AD} = \vec{a}-3\vec{b}+5\vec{c}$. Determinant $= (\lambda-1)(15-12)+2(-10+4)+3(6-3)=0 \Rightarrow 3(\lambda-1)-12+9=0 \Rightarrow \lambda=2$. Answer: 2
Correct Answer: 2