Find the vector and Cartesian equations of the plane passing through the points (2, 2- 1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above.

$\mathbf{OR}$

Find the equation of the plane that contains the lines $\overrightarrow{\mathrm{r}}=\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)+\mathrm{\lambda }\left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)$ and the point (- 1, 3, -4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane, thus obtained.

Given: Assume A(2, 2, -1), B(3, 4, 2) and C(7, 0, 6)

Let $\overrightarrow{\mathrm{a}}=2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}$

$$\begin{gathered} \overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}+4\hat{\mathrm{j}}+2\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{c}}=7\hat{\mathrm{i}}+6\hat{\mathrm{k}} \end{gathered}$$

Hence the vector equation of the plane passing through the points

$$\begin{gathered} \left(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}\right)·\left(\overrightarrow{\mathrm{AB}}\times \overrightarrow{\mathrm{AC}}\right)=0 \\ =\left(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}\right)·\left(\left(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}\right)\times \left(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}\right)\right)=0 \end{gathered}$$

Now,

$$\begin{gathered} \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\left(3\hat{\mathrm{i}}+4\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)-\left(2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right) \\ \Rightarrow \hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}=\left(7\hat{\mathrm{i}}+6\hat{\mathrm{k}}\right)-\left(2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right) \\ =5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+7\hat{\mathrm{k}} \end{gathered}$$

So the required vector equation of plane is

$\left[\overrightarrow{\mathrm{r}}-\left(2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)\right]·\left[\left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)\right]\times \left(5\hat{\mathrm{i}}-2\hat{\mathrm{j}}+7\hat{\mathrm{k}}\right)=0$

Now,

$$\begin{gathered} \left(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\right)\times \left(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}\right)=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 2& 3\\ 5& -2& 7\end{array}\right| \\ =\hat{\mathrm{i}}(14+6)-\hat{\mathrm{i}}(7-15)+\hat{\mathrm{k}}(-2-10) \\ =20\hat{\mathrm{i}}+8\hat{\mathrm{j}}-12\hat{\mathrm{k}} \\ \Rightarrow \left(\overrightarrow{\mathrm{r}}-\left(2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)\right)·\left(20\hat{\mathrm{i}}+8\hat{\mathrm{j}}-12\hat{\mathrm{k}}\right)=0 \\ \left(\overrightarrow{\mathrm{r}}-\left(2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)\right)·\left(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\right)=0 \\ \overrightarrow{\mathrm{r}}·\left(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\right)=\left(2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)·\left(5\hat{\mathrm{i}}+2\hat{\mathrm{i}}-\hat{\mathrm{k}}\right) \\ \overrightarrow{\mathrm{r}}·\left(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\right)=10+4+3 \\ \overrightarrow{\mathrm{r}}·\left(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\right)=17 \end{gathered}$$

Which is the required vector equation of the plane.

Now, the Cartesian equations of the plane passing through the three points is given as below,

5x+2y-3z-17=0

Which is the required cartesian equations of the plane.

Now,

The equation of plane parallel to $5\mathrm{x}+2\mathrm{y}-3\mathrm{z}-17=0$ will be $5\mathrm{x}+2\mathrm{y}-3\mathrm{z}+\mathrm{\lambda }=0$

$\therefore$ it passes through (4, 3, 1)

So, $5\times 4+2\times 3-3\times 1+\mathrm{\lambda }=0$

$20+6-3+\mathrm{\lambda }=0$

So, $\mathrm{\lambda }=-23$

so the equation of the plane will be

$$\begin{gathered} 5\mathrm{x}+2\mathrm{y}-3\mathrm{z}-23=0 \\ 5\mathrm{x}+2\mathrm{y}-3\mathrm{z}=23 \end{gathered}$$

so the vector form of the equation of plane will be

$\overrightarrow{\mathrm{r}}·\left(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\right)=23$

Therefore, the vector and Cartesian equations of the plane passing through the points (2, 2-1), (3, 4, 2) and (7, 0, 6) are $\overrightarrow{\mathrm{r}}·\left(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\right)=17$ and 5x+2y-3z-17=0 respectively and the vector equation of a plane passing through (4, 3, 1) and parallel to the obtained plane is $\overrightarrow{\mathrm{r}}·\left(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\right)=23$.

                                           OR

Let the vector equation of the required plane be $\overrightarrow{\mathrm{r}}·\overrightarrow{\mathrm{n}}=\mathrm{d}$

Given that the plane contains the line $\overrightarrow{\mathrm{r}}=\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)+\mathrm{\lambda }\left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{j}}\right).$

Since the plane passes through point A and B. So $\overrightarrow{\mathrm{n}}$ will be parallel to vector

$$\begin{gathered} \overrightarrow{\mathrm{AB}}\times \left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right) \\ \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ =\left(-\hat{\mathrm{i}}+3\hat{\mathrm{j}}-4\hat{\mathrm{k}}\right)-\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right) \\ =-2\hat{\mathrm{i}}+2\hat{\mathrm{j}}-4\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{AB}}\times \left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 2& -1\\ -2& 2& -4\end{array}\right| \\ =\hat{\mathrm{i}}(-8+2)-\hat{\mathrm{j}}(-4-2)+\hat{\mathrm{k}}(2+4) \\ =-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}+6\hat{\mathrm{k}} \end{gathered}$$

which is a normal vector to the plane.

So the equation of plane will be $\overrightarrow{\mathrm{r}}·\left(-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}+6\hat{\mathrm{k}}\right)=0$

$\overrightarrow{\mathrm{r}}\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)=0$

in Cartesian plane,

$$\begin{gathered} \left(\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}+\mathrm{z}\hat{\mathrm{k}}\right)·\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)=0 \\ \mathrm{x}-\mathrm{y}-\mathrm{z}=0 \end{gathered}$$

So, the perpendicular distance of the plane from the point (2, 1, 4) is

$=\left|\frac{2-1-4}{\sqrt{1^2+{(-1)}^2+{(-1)}^2}}=\left|\frac{-3}{\sqrt{3}}\right|=\sqrt{3} \mathrm{unit}.\right|$

Therefore, the equation of the plane that contains the lines $\overrightarrow{\mathrm{r}}=\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}\right)+\mathrm{\lambda }\left(\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)$ and the point (-1, 3, -4) is $\overrightarrow{\mathrm{r}}\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)=0$ and Cartesian equation is x-y-z=0. And the length of the perpendicular drawn from the point (2, 1, 4) to the obtained plane is $\sqrt{3}$ unit.