Let a= i+2 j+4 k, b= i+ j+4 k, c=2 i+4 j+(^2-1) k be coplanar.

Question

Let $\vec a=\hat i+2\hat j+4\hat k$, $\vec b=\hat i+\lambda\hat j+4\hat k$, $\vec c=2\hat i+4\hat j+(\lambda^2-1)\hat k$ be coplanar. The non-zero vector $\vec a	imes\vec c$ is:

Accepted Answer

$$\begin{vmatrix}1&2&4\1&\lambda&4\2&4&\lambda^2-1\end{vmatrix}=0 \implies \lambda^3-2\lambda^2-9\lambda+18=0 \implies (\lambda-2)(\lambda-3)(\lambda+3)=0$$

For $\lambda=3$ or $-3$: $\vec c=2\vec a \Rightarrow \vec a\times\vec c=\vec 0$ (rejected). For $\lambda=2$: $\vec c=2\hat i+4\hat j+3\hat k$:
$$\vec a\times\vec c = \begin{vmatrix}\hat i&\hat j&\hat k\1&2&4\2&4&3\end{vmatrix} = -10\hat i+5\hat j$$

Solutions · 1