Question

The points A(−1, 3, 0), B(2, 2, 1) and C(1, 1, 3) determine a plane. Find the distance from the plane to the point D(5, 7, 8).
a) \[\sqrt {66} \]
b) \[\sqrt {71} \]
c) \[\sqrt {73} \]
d) \[\sqrt {76} \]

Answer 1

Hint: First find the Cartesian equation of the plane using the three points and then find the perpendicular distance between the plane and the point D.

Formula Used:Cartesian equation of the plane passing through three points \[\left( {{x_1},{y_1},{z_1}} \right),\left( {{x_2},{y_2},{z_2}} \right)\,{\rm{and}}\left( {{x_3},{y_3},{z_3}} \right)\] is
\[\left| {\begin{array}{*{20}{c}}{x - {x_1}}&{y - {y_1}}&{z - {z_1}}\\{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\end{array}} \right| = 0\]



The perpendicular distance of the point \(({x_1},{y_1},{z_1})\)from the plane \(ax + by + cz + d = 0\)is \(d = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)



Complete step by step solution:Cartesian equation of the plane passing through three points A(−1,3,0),
B(2, 2, 1) and C(1,1, 3) is
\[\left| {\begin{array}{*{20}{c}}{x - {x_1}}&{y - {y_1}}&{z - {z_1}}\\{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\end{array}} \right| = 0\]
Here, A(−1,3,0) = \[\left( {{x_1},{y_1},{z_1}} \right)\], B(2, 2, 1) = \[\left( {{x_2},{y_2},{z_2}} \right)\] and C(1, 1, 3) = \[\left( {{x_3},{y_3},{z_3}} \right)\]
                                       \[\left| {\begin{array}{*{20}{c}}{x + 1}&{y - 3}&{z - 0}\\{2 + 1}&{2 - 3}&{1 - 0}\\{1 + 1}&{1 - 3}&{3 - 0}\end{array}} \right| = 0\]
                                             \[\left| {\begin{array}{*{20}{c}}{x + 1}&{y - 3}&z\\3&{ - 1}&1\\2&{ - 2}&3\end{array}} \right| = 0\]
      \[(x + 1)( - 3 + 2) - (y - 3)(9 - 2) + z( - 6 + 2) = 0\]
                       \[(x + 1)( - 1) - (y - 3)(7) + z( - 4) = 0\]
                                       \[ - x - 1 - 7y + 21 - 4z = 0\]
                                              \[x + 7y + 4z - 20 = 0\]
Perpendicular distance of the point D(5, 7, 8) from the plane\[x + 7y + 4z - 20 = 0\]
\( = \left| {\dfrac{{(1)(5) + (7)(7) + (4)(8) - 20}}{{\sqrt {{1^2} + {7^2} + {4^2}} }}} \right|\)
\( = \left| {\dfrac{{5 + 49 + 32 - 20}}{{\sqrt {1 + 49 + 16} }}} \right|\)
\( = \left| {\dfrac{{66}}{{\sqrt {66} }}} \right|\)
\( = \sqrt {66} \)



Option ‘A’ is correct



Note: The student may make a mistake in the expansion of the determinant. The student may have used the incorrect formula for calculating the perpendicular distance from a point to the plane as the perpendicular distance from the origin to the plane.