Question

Find the vector equation of the plane passing through points A(1, 1, 2), B(0, 2, 3) and C(4, 5, 6).

Answer 1

The vector equation of the plane passing through three non-collinear points ${\displaystyle A\left(\overline{a}\right),B\left(\overline{b}\right)}$ and ${\displaystyle C\left(\overline{c}\right)}$ is ${\displaystyle \overline{r}.(\overline{AB}\times \overline{AC})=\overline{a}.(\overline{AB}\times \overline{AC})}$  ...(1)

Here, ${\displaystyle \overline{a}=\hat{i}+\hat{j}+2\hat{k},\overline{b}=2\hat{j}+3\hat{k},\overline{c}=4\hat{i}+5\hat{j}+6\hat{k}}$

∴ ${\displaystyle \overline{AB}=\overline{b}-\overline{a}}$

= ${\displaystyle (2\hat{j}+3\hat{k})-(\hat{i}+\hat{j}+2\hat{k})}$

= ${\displaystyle -\hat{i}+\hat{j}+\hat{k}}$

∴ ${\displaystyle \overline{AC}=\overline{c}-\overline{a}}$

= ${\displaystyle (4\hat{i}+5\hat{j}+6\hat{k})-(\hat{i}+\hat{j}+2\hat{k})}$

= ${\displaystyle 3\hat{i}+4\hat{j}+4\hat{k}}$

∴ ${\displaystyle \overline{AB}\times \overline{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 1& 1\\ 3& 4& 4\end{array}\right|}$

= ${\displaystyle (4-4)\hat{i}-(-4-3)\hat{j}+(-4-3)\hat{k}}$

= ${\displaystyle 0\hat{i}-7\hat{j}+7\hat{k}}$

Now, ${\displaystyle \overline{a}.(\overline{AB}\times \overline{AC})}$

= ${\displaystyle (\hat{i}+\hat{j}+2\hat{k}).(0\hat{i}-7\hat{j}+7\hat{k})}$

= (1)(0) − (1)(7) + (2)(7)

= 0 − 7 + 14

∴ From (1), the vector equation of the required plane is ${\displaystyle \overline{r}.(0\hat{i}-7\hat{j}+14\hat{k})}$ = −7.