The distance of the point (3, 2, -1) from the plane 3x−y+4z+1=0 along the line \[{{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}\] is equal to :
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Official solution
Line PQ is parallel to line \[{{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}\]
\[\therefore\] DR of PQ = DR of line = <-2, 2, 1>
\[\therefore\] Equation of line PQ passing through P(3, 2, -1) and DR = <-2, 2, 1> is
\[{{x - 3} \over { - 2}} = {{y - 2} \over 2} = {{z + 1} \over 1}\]
Any General point on line PQ = (x1, y1, z1)
\[\therefore\] \[{{{x_1} - 3} \over { - 2}} = {{{y_1} - 2} \over 2} = {{{z_1} + 1} \over 1} = \lambda \]
\[ \Rightarrow {x_1} = - 2\lambda + 3\]
\[{y_1} = 2\lambda + 2\]
\[{z_1} = \lambda - 1\]
\[\therefore\] Point Q = (-2\[\lambda\] + 3, 2\[\lambda\] + 2, \[\lambda\] - 1)
Point Q lies on the plane 3x−y+4z+1=0. So point Q satisfy the equation.
\[3( - 2\lambda + 3) - (2\lambda + 2) + 4(\lambda - 1) + 1 = 0\]
\[ \Rightarrow - 6\lambda + 9 - 2\lambda - 2 + 4\lambda - 4 + 1 = 0\]
\[ \Rightarrow - 4\lambda + 4 = 0\]
\[ \Rightarrow \lambda = 1\]
\[\therefore\] Point Q = (-2 \[\times\] 1 + 3, 2 \[\times\] 1 + 2, 1 - 1)
= (1, 4, 0)
\[\therefore\] Distance of the point P(3, 2, -1) from the plane = Length of PQ
\[ = \sqrt {{{(3 - 1)}^2} + {{(2 - 4)}^2} + {{( - 1 - 0)}^2}} \]
\[ = \sqrt {{2^2} + {{( - 2)}^2} + {{( - 1)}^2}} \]
\[ = \sqrt {4 + 4 + 1} \]
\[ = \sqrt 9 \]
=3