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

 

Answer 1

The equations of the given lines are  \[\vec{r} = \left( \hat{i}  + 2 \hat{j} - 4 \hat{k}  \right) + \lambda\left( 2 \hat{i}  + 3 \hat{j}  + 6 \hat{k} \right)\] \[\vec{r} = \left( \hat{i}  - 3 \hat{j} + 5 \hat{k}  \right) + \mu\left( \hat{i}  + \hat{j} - \hat{k}  \right)\] We know that the vector equation of a plane passing through a point  \[\vec{a}\] and parallel to  \[\vec{b}\] and  \[\vec{c}\]  is given by \[\left( \vec{r} - \vec{a} \right) . \left( \vec{b} \times \vec{c} \right) = 0\]  

Here,  \[\vec{a} = \hat{i}+ 2 \hat{j}  - 4 \hat{k} \]

\[\vec{b} = 2 \hat{i} + 3 \hat{j}  + 6 \hat{k} \] and 

\[\vec{c} = \hat{i}  + \hat{j}  - \hat{k} \]

\[\therefore \vec{b} \times \vec{c} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 2 & 3 & 6 \\ 1 & 1 & - 1\end{vmatrix} = - 9 \hat{i} + 8 \hat{j}  - \hat{k} \]

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

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

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  \[- 9x + 8y - z - 11 = 0\]

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