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प्रश्न
Find the foot of the perpendicular drawn from the point \[\hat{i} + 6 \hat{j} + 3 \hat{k} \] to the line \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) .\] Also, find the length of the perpendicular
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उत्तर
Let L be the foot of the perpendicular drawn from the point P ( \[\hat{i} + 6 \hat{j} + 3 \hat{k}\] ) to the line \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) .\]
Let the position vector L be \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = \lambda \hat{i} + \left( 1 + 2\lambda \right) \hat{j} + \left( 2 + 3\lambda \right) \hat{k} \]
Now,
\[\overrightarrow{PL} = \text{ Position vector of L - Position vector of P } \]
\[ \Rightarrow \overrightarrow{PL} = \left\{ \lambda \hat{i} + \left( 1 + 2\lambda \right) \hat{j} + \left( 2 + 3\lambda \right) \hat{k} \right\} - \left( \hat{i} + 6 \hat{j} + 3 \hat{k} \right)\]
\[ \Rightarrow \overrightarrow{PL} = \left( \lambda - 1 \right) \hat{i} + \left( 2\lambda - 5 \right) \hat{j} + \left( 3\lambda - 1 \right) \hat{k} . . . (2)\]
Since
\[\overrightarrow{PL}\] is perpendicular to the given line, which is parallel to \[\overrightarrow{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
we have ,
\[\overrightarrow{PL} . \overrightarrow{b} = 0\]
\[ \Rightarrow \left\{ \left( \lambda - 1 \right) \hat{i} + \left( 2\lambda - 5 \right) \hat{j} + \left( 3\lambda - 1 \right) \hat{k} \right\} . \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 0 \]
\[ \Rightarrow 1\left( \lambda - 1 \right) + 2\left( 2\lambda - 5 \right) + 3\left( 3\lambda - 1 \right) = 0\]
\[ \Rightarrow \lambda = 1\]
Substituting \[\lambda = 1\] in (1),
we get the position vector of L as \[\hat{i} + 3 \hat{j} + 5 \hat{k}\]
Substituting \[\lambda = 1\] in (2),
we get , \[\overrightarrow{PL} = - 3 \hat{j} + 2 \hat{k}\]
\[= \sqrt{\left( - 3 \right)^2 + 2^2}\]
\[ = \sqrt{13}\]
Hence, the length of the perpendicular from point P on PL is \[\sqrt{13} \text{ units} \] .
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