Question

Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines

$\overrightarrow{r}=\left(\hat{i}+2\hat{j}-4\hat{k}\right)+\lambda \left(2\hat{i}+3\hat{j}+6\hat{k}\right)$ and $\overrightarrow{r}=\left(\hat{i}-3\hat{j}+5\hat{k}\right)+\mu \left(\hat{i}+\hat{j}-\hat{k}\right)$. Also, find the distance of the point (9, −8, −10) from the plane thus obtained. [CBSE 2014]

Answer 1

The equations of the given lines are

$\overrightarrow{r}=\left(\hat{i}+2\hat{j}-4\hat{k}\right)+\lambda \left(2\hat{i}+3\hat{j}+6\hat{k}\right)$

$\overrightarrow{r}=\left(\hat{i}-3\hat{j}+5\hat{k}\right)+\mu \left(\hat{i}+\hat{j}-\hat{k}\right)$

We know that the vector equation of a plane passing through a point $\overrightarrow{a}$ and parallel to $\overrightarrow{b}$ and $\overrightarrow{c}$ is given by $\left(\overrightarrow{r}-\overrightarrow{a}\right).\left(\overrightarrow{b}\times \overrightarrow{c}\right)=0$.

Here, $\overrightarrow{a}=\hat{i}+2\hat{j}-4\hat{k}$, $\overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k}$ and $\overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$.

$\therefore \overrightarrow{b}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 6\\ 1& 1& -1\end{array}\right|=-9\hat{i}+8\hat{j}-\hat{k}$

So, the vector equation of the plane is

$$\begin{gathered} \left(\overrightarrow{r}-\overrightarrow{a}\right).\left(\overrightarrow{b}\times \overrightarrow{c}\right)=0 \\ \Rightarrow \left[\overrightarrow{r}-\left(\hat{i}+2\hat{j}-4\hat{k}\right)\right].\left(-9\hat{i}+8\hat{j}-\hat{k}\right)=0 \\ \Rightarrow \overrightarrow{r}.\left(-9\hat{i}+8\hat{j}-\hat{k}\right)=\left(\hat{i}+2\hat{j}-4\hat{k}\right).\left(-9\hat{i}+8\hat{j}-\hat{k}\right) \\ \Rightarrow \overrightarrow{r}.\left(-9\hat{i}+8\hat{j}-\hat{k}\right)=1\times \left(-9\right)+2\times 8+\left(-4\right)\times \left(-1\right)=-9+16+4=11 \end{gathered}$$

Thus, the vector equation of the plane is $\overrightarrow{r}.\left(-9\hat{i}+8\hat{j}-\hat{k}\right)=11$.

The Cartesian equation of this plane is

$$\begin{gathered} \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(-9\hat{i}+8\hat{j}-\hat{k}\right)=11 \\ \Rightarrow -9x+8y-z=11 \end{gathered}$$

Now,

Distance of the point (9, −8, −10) from the plane $-9x+8y-z=11$

= Length of perpendicular from (9, −8, −10) from the plane $-9x+8y-z-11=0$

$$\begin{gathered} =\left|\frac{-9\times 9+8\times \left(-8\right)-\left(-10\right)-11}{\sqrt{{\left(-9\right)}^2+8^2+{\left(-1\right)}^2}}\right| \\ =\left|\frac{-81-64+10-11}{\sqrt{81+64+1}}\right| \\ =\left|\frac{-146}{\sqrt{146}}\right| \\ =\sqrt{146}\mathrm{units} \end{gathered}$$