Question

Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, -3), B(-2, -3, 5) and C(5, 3, -3).

Answer 1

To find the distance between the point and plane we first find the equation of plane using the 3 points. Then we use the distance formula from the point to the plane to find the answer.

Complete step by step answer:
Given Data – Points A, B and C compute the plane.

The general equation of the plane is
${\text{a}}\left( {{\text{x - }}{{\text{x}}_1}} \right) + {\text{b}}\left( {{\text{y - }}{{\text{y}}_1}} \right) + {\text{c}}\left( {{\text{z - }}{{\text{z}}_1}} \right) = 0{\text{ where }}\left( {{{\text{x}}_1},{{\text{y}}_1},{{\text{z}}_1}} \right){\text{ is the point}}$
Hence, the general equation of the plane passing through A (2, 5, −3) is
${\text{a}}\left( {{\text{x - 2}}} \right) + {\text{b}}\left( {{\text{y - 5}}} \right) + {\text{c}}\left( {{\text{z + 3}}} \right) = 0{\text{ - - - - - (1)}}$
Now, this plane also passes through point B (−2,−3,5)
Therefore the plane equation should satisfy this point B, let us substitute point B in the plane equation given by equation (1), we get
${\text{a}}\left( {{\text{ - 2 - 2}}} \right) + {\text{b}}\left( {{\text{ - 3 - 5}}} \right) + {\text{c}}\left( {{\text{5 + 3}}} \right) = 0 \\\ \Rightarrow {\text{ - 4a - 8b + 8c = 0}} \\\ \Rightarrow {\text{ - a - 2b + 2c = 0 - - - - - - (2)}} \\\$
Now, this plane also passes through point C(5,3,-3)
Therefore the plane equation should satisfy this point C, let us substitute point C in the plane equation given by equation (1), we get
${\text{a}}\left( {{\text{5 - 2}}} \right) + {\text{b}}\left( {{\text{3 - 5}}} \right) + {\text{c}}\left( {{\text{ - 3 + 3}}} \right) = 0 \\\ \Rightarrow 3{\text{a - 2b = 0 - - - - - - (3)}} \\\$
Solving (2) and (3), we get
${\text{ - a - 2b + 2c = 0}}$ and $3{\text{a - 2b = 0}}$
⟹3a = -a + 2c
⟹4a = 2c
⟹4a = 2c, 2b = 3a, 8b = 12a
$\Rightarrow \dfrac{{\text{a}}}{4} = \dfrac{{\text{b}}}{6} = \dfrac{{\text{c}}}{8} \\\ {\text{let }}\dfrac{{\text{a}}}{4} = \dfrac{{\text{b}}}{6} = \dfrac{{\text{c}}}{8} = \lambda \\\ \Rightarrow {\text{a = 4}}\lambda {\text{, b = 6}}\lambda {\text{, c = 8}}\lambda \\\ \\\$
Substituting the values of a, b and c in (1), we get
$\Rightarrow 4\lambda \left( {{\text{x - 2}}} \right) + 6\lambda \left( {{\text{y - 5}}} \right) + 8\lambda \left( {{\text{z + 3}}} \right) = 0 \\\ \Rightarrow 4\left( {{\text{x - 2}}} \right) + 6\left( {{\text{y - 5}}} \right) + 8\left( {{\text{z + 3}}} \right) = 0 \\\ \Rightarrow {\text{4x - 8 + 6y - 30 + 8z + 24 = 0}} \\\ \Rightarrow {\text{4x + 6y + 8z - 14 = 0}} \\\ \Rightarrow {\text{2x + 3y + 4z - 7 = 0}} \\\$
Hence, the equation of the plane determined by the points A(2,5,−3),B(−2,−3,5) and C(5,3,−3) is 2x+3y+4z−7 = 0
Distance from a point $\left( {{{\text{x}}_1},{{\text{y}}_1},{{\text{z}}_1}} \right)$to a plane whose equation is ax+by+cz+d=0 is given by D = $\dfrac{{|{\text{a}}{{\text{x}}_1} + {\text{b}}{{\text{y}}_1} + {\text{c}}{{\text{z}}_1} + {\text{d|}}}}{{\sqrt {{{\text{a}}^2} + {{\text{b}}^2} + {{\text{c}}^2}} }}$
Now, the distance (D) of this plane from the point (7, 2, 4) is
${\text{ = }}\dfrac{{|2 \times 7 + 3 \times 2 + 4 \times 4 - 7{\text{|}}}}{{\sqrt {{2^2} + {3^2} + {4^2}} }} \\\ = \dfrac{{|14 + 6 + 16 - 7{\text{|}}}}{{\sqrt {4 + 9 + 16} }} \\\ = \dfrac{{|29{\text{|}}}}{{\sqrt {29} }} \\\ = \sqrt {29} {\text{units}} \\\$

Thus, the distance between the point (7, 2, 4) and the plane 2x+3y+4z−7 = 0 is $\sqrt {29}$ units.

Note: In order to solve this type of problems the key is to have adequate knowledge in concepts of planes and lines in geometry. The distance between a point and a plane is the distance between the point and the normal vector passing through the plane. Any value bounded by the modulus function returns a positive value only.