Let a, b, c be non-zero, non-coplanar vectors.

Question

Let $\vec a,\vec b,\vec c$ be non-zero, non-coplanar vectors. The position vectors of $A,B,C,D$ are $\vec a-\vec b+\vec c$, $\lambda\vec a-3\vec b+4\vec c$, $-\vec a+2\vec b-3\vec c$, $2\vec a-4\vec b+6\vec c$. If $\vec{AB},\vec{AC},\vec{AD}$ are coplanar, $\lambda$ is:

Answer: 2

Accepted Answer

$\vec{AB}=(\lambda-1)\vec a-2\vec b+3\vec c$, $\vec{AC}=-2\vec a+3\vec b-4\vec c$, $\vec{AD}=\vec a-3\vec b+5\vec c$.

Coplanarity requires the determinant of coefficients to vanish:
$$\begin{vmatrix}\lambda-1&-2&3\-2&3&-4\1&-3&5\end{vmatrix}=0$$
$$(\lambda-1)(15-12)+2(-10+4)+3(6-3)=0 \implies 3(\lambda-1)-12+9=0 \implies 3(\lambda-1)=3 \implies \lambda=2$$

Solutions · 1