Question

The position vectors of the points A, B, C are respectively (1, 1, 1); (1, -1, 2); (0, 2, -1) Find a unit vector parallel to the plane determined by ABC & perpendicular to the vector (1, 0, 1)

• A. $\pm \frac{1}{3\sqrt{3}}(i+5j-k)$
• B. $\pm (i+5j-k)$
• C. $\pm (-i-5j-k)$
• D. $\pm \frac{1}{3\sqrt{3}}(-i-5j-k)$

Answer 1

The correct option is A $\pm \frac{1}{3\sqrt{3}}(i+5j-k)$

Given $\overline{OA}=\hat{i}+\hat{j}+\hat{k};\overline{OB}=\hat{i}-\hat{j}+2\hat{k}$ and $\overline{OC}=2\hat{j}-\hat{k}$
$\overline{AB}=-2\hat{j}+\hat{k}$ and $\overline{AC}=-\hat{i}+\hat{j}-2\hat{k}$
$\overline{AB}\times \overline{AC}=3\hat{i}-\hat{j}-2\hat{k}$
let the required unit vector be $\overline{r}=x\hat{i}+y\hat{j}+z\hat{k}$
As $\overline{r}$ is parallel to the plane determined by $ABC$.
$\overline{r}$ is perpendicular to $\overline{AB}\times \overline{AC}$
$\Rightarrow 3x-y-2z=0$ -----(1)
given $\overline{r}$ is also perpendicular to $\hat{i}+\hat{k}$
$\Rightarrow x+z=0$ -----(2)
$\overline{r}$ is an unit vector.
$\Rightarrow x^2+y^2+z^2=1$ ----(3)
solving equations (1),(2) and (3) gives
$x=\pm \frac{1}{3\sqrt{3}};y=\pm \frac{5}{3\sqrt{3}}$ and $z=\mp \frac{1}{3\sqrt{3}}$
$\therefore \overline{r}=\pm \frac{1}{3\sqrt{3}}(\hat{i}+5\hat{j}-\hat{k})$
Hence, option A.