Question

If the distance of the point \[P\left( {1, - 2,1} \right)\] from the plane \[x + 2y - 2z = \alpha \], where \[\alpha > 0\] is 5, then the foot of the perpendicular from \[P\] is
A. \[\left( {\dfrac{8}{3},\dfrac{4}{3}, - \dfrac{7}{3}} \right)\]
B. \[\left( {\dfrac{4}{3}, - \dfrac{4}{3},\dfrac{1}{3}} \right)\]
C. \[\left( {\dfrac{1}{3},\dfrac{2}{3},\dfrac{{10}}{3}} \right)\]
D. \[\left( {\dfrac{2}{3}, - \dfrac{1}{3},\dfrac{5}{2}} \right)\]

Answer 1

Hint: First of all, find an equation with \[\alpha \] using given data and find its values. Since \[\alpha \] is a positive value, eliminate the negative value of \[\alpha \]. Then use the formula for the foot of the perpendicular to get the required answer.

Complete step-by-step answer:
Given that the distance from \[P\left( {1, - 2,1} \right)\] to the plane \[x + 2y - 2z = \alpha \] is 5.
So, we have
\[
   \Rightarrow \left| {\dfrac{{1 + 2\left( { - 2} \right) - 2\left( 1 \right) - \alpha }}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} }}} \right| = 5 \\
   \Rightarrow \left| {\dfrac{{1 - 4 - 2 - \alpha }}{{\sqrt {1 + 4 + 4} }}} \right| = 5 \\
   \Rightarrow \left| {\dfrac{{ - \left( {5 + \alpha } \right)}}{{\sqrt 9 }}} \right| = 5 \\
   \Rightarrow \left| {\dfrac{{\alpha + 5}}{3}} \right| = 5 \\
   \Rightarrow \dfrac{{\alpha + 5}}{3} = \pm 5 \\
   \Rightarrow \alpha + 5 = \pm 15 \\
   \Rightarrow \alpha = - 5 \pm 15 \\
   \Rightarrow \alpha = - 5 + 15, - 5 - 15 \\
  \therefore \alpha = 10, - 20 \\
\]
Since \[\alpha > 0\], we have \[\alpha = 10\]. Therefore, the equation of the plane is \[x + 2y - 2z = 10\].
Let \[Q\left( {{x_1},{y_1},{z_1}} \right)\] be the foot of the perpendicular from \[P\left( {1, - 2,1} \right)\] to the plane \[x + 2y - 2z = 10\].
So, we have
\[
   \Rightarrow \dfrac{{{x_1} - 1}}{1} = \dfrac{{{y_1} - \left( { - 2} \right)}}{2} = \dfrac{{{z_1} - 1}}{{ - 2}} = \dfrac{{ - \left[ {\left( 1 \right) + 2\left( { - 1} \right) - 2\left( 1 \right) - 10} \right]}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} }} \\
   \Rightarrow \dfrac{{{x_1} - 1}}{1} = \dfrac{{{y_1} + 2}}{2} = \dfrac{{{z_1} - 1}}{{ - 2}} = \dfrac{{ - \left[ {1 - 4 - 2 - 10} \right]}}{{\sqrt {1 + 4 + 4} }} \\
   \Rightarrow \dfrac{{{x_1} - 1}}{1} = \dfrac{{{y_1} + 2}}{2} = \dfrac{{{z_1} - 1}}{{ - 2}} = \dfrac{{15}}{9} \\
   \Rightarrow \dfrac{{{x_1} - 1}}{1} = \dfrac{{{y_1} + 2}}{2} = \dfrac{{{z_1} - 1}}{{ - 2}} = \dfrac{5}{3} \\
\]
Taking first and last terms, we get
\[
   \Rightarrow \dfrac{{{x_1} - 1}}{1} = \dfrac{5}{3} \\
   \Rightarrow 3{x_1} - 3 = 5 \\
   \Rightarrow 3{x_1} = 5 + 3 = 8 \\
  \therefore {x_1} = \dfrac{8}{3} \\
\]
Taking second and last terms, we get
\[
   \Rightarrow \dfrac{{{y_1} + 2}}{2} = \dfrac{5}{3} \\
   \Rightarrow 3{y_1} + 6 = 10 \\
   \Rightarrow 3{y_1} = 10 - 6 = 4 \\
  \therefore {y_1} = \dfrac{4}{3} \\
\]
Taking third and last terms, we get
\[
   \Rightarrow \dfrac{{{z_1} - 1}}{{ - 2}} = \dfrac{5}{3} \\
   \Rightarrow 3{z_1} - 3 = - 10 \\
   \Rightarrow 3{z_1} = - 10 + 3 = - 7 \\
  \therefore {z_1} = \dfrac{{ - 7}}{3} \\
\]
Therefore, the foot of the perpendicular is \[\left( {\dfrac{8}{3},\dfrac{4}{3}, - \dfrac{7}{3}} \right)\].
Thus, the correct option is A. \[\left( {\dfrac{8}{3},\dfrac{4}{3}, - \dfrac{7}{3}} \right)\]

Note: If \[Q\left( {{x_1},{y_1},{z_1}} \right)\] is the foot of perpendicular from the point \[P\left( {{x_2},{y_2},{z_2}} \right)\] to the plane \[ax + by + cz + d = 0\] then we have \[\dfrac{{{x_1} - {x_2}}}{a} = \dfrac{{{y_1} - {y_2}}}{b} = \dfrac{{{z_1} - {z_2}}}{c} = \dfrac{{ - \left( {a{x_2} + b{y_2} + c{z_2} + d} \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\].