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

 

Answer 1

The equations of the given lines are  $$\overrightarrow{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$$ $$\overrightarrow{r}=(\hat{i}-3\hat{j}+5\hat{k})+\mu (\hat{i}+\hat{j}-\hat{k})$$ 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 $$(\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{b}\times \overrightarrow{c})=0$$  

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 $$(\overrightarrow{r}-\overrightarrow{a}).(\overrightarrow{b}\times \overrightarrow{c})=0$$
$$\Rightarrow [\overrightarrow{r}-(\hat{i}+2\hat{j}-4\hat{k})].(-9\hat{i}+8\hat{j}-\hat{k})=0$$
$$\Rightarrow \overrightarrow{r}.(-9\hat{i}+8\hat{j}-\hat{k})=(\hat{i}+2\hat{j}-4\hat{k}).(-9\hat{i}+8\hat{j}-\hat{k})$$
$$\Rightarrow \overrightarrow{r}.(-9\hat{i}+8\hat{j}-\hat{k})=1\times (-9)+2\times 8+(-4)\times (-1)=-9+16+4=11$$ 
Thus, the vector equation of the plane is

$$\overrightarrow{r}.(-9\hat{i}+8\hat{j}-\hat{k})=11$$ The Cartesian equation of this plane is $$(x\hat{i}+y\hat{j}+z\hat{k}).(-9\hat{i}+8\hat{j}-\hat{k})=11$$
$$\Rightarrow -9x+8y-z=11$$

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  $$-9x+8y-z-11=0$$

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