Question

Find the distance of the point (3, 3, 3) from the plane \[\vec{r} \cdot \left( 5 \hat{i}  + 2 \hat{j}  - 7k \right) + 9 = 0\]

 

Answer 1

\[\text{ The given plane is } \]
\[ \vec{r} . \left( 5 \hat{i}  + 2 \hat{j}  - 7 \hat{k}  \right) + 9 = 0\]
\[ \Rightarrow \vec{r} . \left( - 5 \hat{i}  - 2 \hat{j}  + 7 \hat{k} \right) = 9\]
\[ \text{ We know that the perpendicular distance of a point P of position vector }  \vec{a} \text{ from the plane } \vec{r} . \vec{n} =\text{ d is given by }  \]
\[p = \frac{\left| \vec{a} . \vec{n} - d \right|}{\left| \vec{n} \right|}\]
\[ \text{ Finding the distance from (3, 3, 3) (which means 3 }  \hat{i}  +3 \hat{j}  + 3 \hat{k}  ) \text{ to the given plane } \]
\[\text{ Here } , \vec{a} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} ; \vec{n} = - 5 \hat{i} - 2 \hat{j} + 7 \hat{k}  ; d = 9\]
\[ \text{ So, the required distance p } \]
\[ = \frac{\left| \left( 3 \hat{i} + 3 \hat{j} + 3 \hat{k}  \right) . \left( - 5 \hat{i}  - 2 \hat{j}  + 7 \hat{k} \right) - 9 \right|}{\left| 3 \hat{i} + 3 \hat{j}  + 3 \hat{k}  \right|}\]
\[ = \frac{\left| - 15 - 6 + 21 - 9 \right|}{\sqrt{25 + 4 + 49}}\]
\[ = \frac{\left| - 9 \right|}{\sqrt{78}}\]
\[ = \frac{9}{\sqrt{78}} \text{ units } \]