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Question
Show that the points \[\hat{i} - \hat{j} + 3 \hat{k} \text{ and } 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \] are equidistant from the plane \[\vec{r} \cdot \left( 5 \hat{i} + 2 \hat{j} - 7 \hat{k} \right) + 9 = 0 .\]
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Solution
\[\text{ The given plane is } \]
\[ \vec{r} . \left( 5 \hat{i} + 2 \hat{j} - 7 \hat{k} \right) + 9 = 0\]
\[ \Rightarrow \vec{r} . \left( - 5 \hat{i} - 2 \hat{j} + 7 \hat{k} \right) = 9\]
\[ \text{ We know that the perpendicular distance of a point P of position vector } \vec{a} \text{ from the plane } \vec{r} . \vec{n} = \text{ d is given by} \]
\[p = \frac{\left| \vec{a} . \vec{n} - d \right|}{\left| \vec{n} \right|}\]
\[ \text{ Finding the distance from } \hat{i} - \hat{j} + 3 \hat{k} \text{ to the given plane } \]
\[ \text{ Here } , \vec{a} = \hat{i} - \hat{j} + 3 \hat{k} ; \vec{n} = - 5 \hat{i} - 2 \hat{j} + 7 \hat{k} ; d = 9\]
\[ \text{ So, the required distance p is given by } \]
\[p = \frac{\left| \left( \hat{i} - \hat{j} + 3 \hat{k} \right) . \left( - 5 \hat{i} - 2 \hat{j} + 7 \hat{k} \right) - 9 \right|}{\left| - 5 \hat{i} - 2 \hat{j} + 7 \hat{k} \right|}\]
\[ = \frac{\left| - 5 + 2 + 21 - 9 \right|}{\sqrt{25 + 4 + 49}}\]
\[ = \frac{\left| 9 \right|}{\sqrt{78}}\]
\[ = \frac{9}{\sqrt{78}} \text{ units } ... (1)\]
\[ \text{ Finding the distance from } 3 \hat{i} +3 \hat{j} + 3 \hat{k} \text{ to the given plane } \]
\[ \text{ Here } , \vec{a} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} ; \vec{n} = - 5 \hat{i} - 2 \hat{j} + 7 \hat{k} ; d = 9\]
\[ \text{ So, the required distance p is given by } \]
\[p = \frac{\left| \left( 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \right) . \left( - 5 \hat{i} - 2 \hat{j} + 7 \hat{k} \right) - 9 \right|}{\left| 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \right|}\]
\[ = \frac{\left| - 15 - 6 + 21 - 9 \right|}{\sqrt{25 + 4 + 49}}\]
\[ = \frac{\left| - 9 \right|}{\sqrt{78}}\]
\[ = \frac{9}{\sqrt{78}} \text{ units } ... (2)\]
\[ \text{ From (1) and (2), we can say that the given points are equidistant from the given plane } .\]
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