If the foot of the perpendicular from the point \[\mathrm{A}(-1,4,3)\] on the plane \[\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4\], is \[\left(-2, \frac{7}{2}, \frac{3}{2}\right)\], then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3,−1,−4, is equal to :
Solutions · 1
Official solution
\[\left( { - 2,{7 \over 2},{3 \over 2}} \right)\] satisfies the plane P:2x+my+nz=4
\[ - 4 + {{7m} \over 2} + {{3n} \over 2} = 4 \Rightarrow 7m + 3n = 16\] ...... (i)
Line joining \[A( - 1,\,4,\,3)\] and \[\left( { - 2,{7 \over 2},{3 \over 2}} \right)\] is perpendicular to P:2x+my+nz=4
\[{1 \over 2} = {{{1 \over 2}} \over m} = {{{3 \over 2}} \over n} \Rightarrow m = 1\] \& n=3
Plane P:2x+y+3z=4
Distance of P from \[A( - 1,\,4,\,3)\] parallel to the line
\[{{x + 1} \over 3} = {{y - 4} \over { - 1}} = {{z - 3} \over { - 4}}:L\]
for point of intersection of P \& L
\[2(3r - 1) + ( - r + 4) + 3( - 4r + 3) = 4 \Rightarrow r = 1\]
Point of intersection : \[(2,\,3,\, - 1)\]
Required distance \[ = \sqrt {{3^2} + {1^2} + {4^2}} = \sqrt {26} \]