Question

Find the equation of a plane passing through the point P(6,5,9) and parallel to the plane determined by the points A(3,-1,2), B(5,2,4) and C(-1,-1,6). Also find the distance of this plane from the point A.

Answer 1

Let $\overrightarrow{AB}=5\hat{i}+2\hat{j}+4\hat{k}-3\hat{i}+\hat{j}-2\hat{k}=2\hat{i}+3\hat{j}+2\hat{k},\overrightarrow{BC}=-\hat{i}-\hat{j}+6\hat{k}-5\hat{i}-2\hat{j}+6\hat{k}-5\hat{i}-2\hat{j}-4\hat{k}=-6\hat{i}-3\hat{j}+2\hat{k}$

So, normal to the plane determined by the points A,B and C is $\overrightarrow{m}=\overrightarrow{AB}\times \overrightarrow{BC}$

$\therefore \overrightarrow{m}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 2\\ -6& -3& 2\end{array}\right|=12\hat{i}-16\hat{j}+12\hat{k}\therefore$ d.r.'s of normal to the required plane $\pi$ : 3,-4,3

[Note that the required plane and plane through A, B, C are parallel.]

Also the plane $\pi$ passes through P(6,5,9) i.e., $\overrightarrow{a}=6\hat{i}+5\hat{j}+9\hat{k}$

$\therefore \pi :\overrightarrow{r}.\overrightarrow{m}=\overrightarrow{a}.\overrightarrow{m}\Rightarrow \overrightarrow{r}.$ $(3\hat{i}-4\hat{j}+3\hat{k})=(6\hat{i}+5\hat{j}+9\hat{k}).(3\hat{3}-4\hat{j}+3\hat{k})$

$\Rightarrow \overrightarrow{r}.(3\hat{i}-4\hat{j}+3\hat{k})$ = 18-20+27

$\therefore \overrightarrow{r}.(3\hat{i}-4j+3\hat{k})==25or3x-4y+3z=25...\left(i\right)$

Now distance of (i) from A(3,-1,2) = $\frac{|3\times 3-4\times (-1)+3\times 2-25|}{\sqrt{3^2+(-4{)}^2+3^2}}=\frac{6}{\sqrt{34}}$ units.