Question 356 of 521MediumMCQ
The length of the perpendicular from the point (2,−1,4) on the straight line 10x+3=−7y−2=1z is:
Solutions · 1
Official solution
Let \[{{x + 3} \over {10}}\]= \[{{y - 2} \over {-7}}\] = \[{{z} \over {1}}\] = \[\lambda \]
\[ \therefore \] General point Q(10\[\lambda \] - 3, -7\[\lambda \] + 2, \[\lambda \]).
\[\overrightarrow {PQ} \] = (10\[\lambda \] - 5) (\widehat i\] + (3 - 7\[\lambda \]) (\widehat j\] + (\[\lambda \] - 4) (\widehat k\]
The vector parrallel to the given line is
= 10\[\widehat i\] - 7\[\widehat j\] + \[\widehat k\]
As \[\overrightarrow {PQ} \] and (10\[\widehat i\] - 7\[\widehat j\] + \[\widehat k\]) are perpendicular to the each other.
\[\overrightarrow {PQ} \] . (10\[\widehat i\] - 7\[\widehat j\] + \[\widehat k\]) = 0
\[ \Rightarrow \] (10\[\lambda \] - 5)10 + (3 - 7\[\lambda \])(-7) + (\[\lambda \] - 4)(1) = 0
\[ \Rightarrow \] 100\[\lambda \] -- 50 + 49\[\lambda \] -- 21 + \[\lambda \] -- 4 = 0
\[ \Rightarrow \] 150\[\lambda \] = 75
\[ \Rightarrow \] \[\lambda \] = \[{1 \over 2}\]
\[ \therefore \] The point Q = (2, \[ - {3 \over 2}\], \[{1 \over 2}\])
\[ \therefore \] Length of perpendicular (PQ) = \[\sqrt {0 + {1 \over 4} + {{49} \over 4}} \]
= \[\sqrt {{{50} \over 4}} = {5 \over {\sqrt 2 }}\] = 3.53
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