Let P be a plane passing through the points (1,0,1), (1,-2,1) and (0,1,-2).

Question

Let $P$ be a plane passing through the points $(1,0,1)$, $(1,-2,1)$ and $(0,1,-2)$. Let a vector $\vec a=\alpha\hat i+\beta\hat j+\gamma\hat k$ be such that $\vec a$ is parallel to the plane $P$, perpendicular to $\hat i+2\hat j+3\hat k$, and $\vec a\cdot(\hat i+\hat j+2\hat k)=2$. Then $(\alpha-\beta+\gamma)^2$ equals \_\_\_\_\_\_.

Answer: 81

Accepted Answer

Normal to $P$: using $v_1=(0,-2,0)$, $v_2=(-1,1,-3)$,
$$\vec n_P=v_1\times v_2=(6,0,-2)\parallel(3,0,-1).$$
$\vec a$ parallel to $P\Rightarrow \vec a\cdot\vec n_P=0\Rightarrow 3\alpha-\gamma=0\Rightarrow\gamma=3\alpha$.
$\vec a\perp(1,2,3)\Rightarrow \alpha+2\beta+3\gamma=0\Rightarrow \alpha+2\beta+9\alpha=0\Rightarrow\beta=-5\alpha$.
$\vec a\cdot(1,1,2)=2\Rightarrow \alpha+\beta+2\gamma=2\Rightarrow \alpha-5\alpha+6\alpha=2\Rightarrow\alpha=1$.
So $\alpha=1,\beta=-5,\gamma=3$, and $(\alpha-\beta+\gamma)^2=(1+5+3)^2=81$.

Solutions · 1