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Question
Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane \[\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 4 \hat{k} \right) + 5 = 0 .\]
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Solution
\[\text{ Let M be the foot of the perpendicular of the point P (1, 1, 2) in the plane } \vec{r} .\left( \hat{i} - 2 \hat{j} + 4 \hat{k} \right)+ 5 = 0 \text{ or } x - 2y + 4z + 5 = 0 . \]
\[\text{ Then,PM is the normal to the plane. So, the direction ratios of PM are proportional to 1, -2, 4. } \]
\[\text{ Since PM passes through P (1, 1, 2) and has direction ratios proportional to 1, - 2, 4 equation of PQ is} \]
\[\frac{x - 1}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{4} = r (\text{ say } )\]
\[\text{ Let the coordinates of M be } \left( r + 1, - 2r + 1, 4r + 2 \right).\]
\[\text{ Since M lies in the plane x }- 2y + 4z + 5 = 0, \]
\[x - 2y + 4z + 5 = 0\]
\[ \Rightarrow r + 1 + 4r - 2 + 16r + 8 + 5 = 0\]
\[ \Rightarrow 21r + 12 = 0\]
\[ \Rightarrow r = \frac{- 12}{21} = \frac{- 4}{7}\]
\[\text{ Substituting this in the coordinates of M, we get } \]
\[M = \left( r + 1, - 2r + 1, 4r + 2 \right) = \left( \left( \frac{- 4}{7} \right) + 1, - 2 \left( \frac{- 4}{7} \right) + 1, 4 \left( \frac{- 4}{7} \right) + 2 \right) = \left( \frac{3}{7}, \frac{15}{7}, \frac{- 2}{7} \right)\]
\[\text{ Now, the length of the perpendicular from P onto the given plane } \]
\[ = \frac{\left| \left( 1 \right) - 2 \left( 1 \right) + 4 \left( 2 \right) + 5 \right|}{\sqrt{1 + 4 + 16}}\]
\[ = \frac{12}{\sqrt{21}} \text{ units } \]
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