
What is the distance of the point \[\left( {2,3,4} \right)\] from the plane \[3x - 6y + 2x + 11 = 0\]?
A. 1
B. 3
C. 2
D. 0
Answer
163.8k+ views
Hint: Here, an equation of a plane and a point is given. by using the formula of the distance between the point and a plane calculate the distance between the given point and the required plane. In the end, simplify the equation of the plane and get the required answer.
Formula Used:The smallest distance between a point \[\left( {{x_1},{y_1},{z_1}} \right)\] and a plane \[ax + by + cz = d\] 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:The given point is \[\left( {2,3,4} \right)\] and the equation of the plane is \[3x - 6y + 2x + 11 = 0\].
By comparing the given information with the formula of the distance between the plane and a point, we get
\[{x_1} = 2,{y_1} = 3,{z_1} = 4\] and \[a = 3,b = - 6,c = 2\].
Now substitute the values in the formula of the distance between the plane and a point \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
\[D = \left| {\dfrac{{\left( 3 \right)\left( 2 \right) + \left( { - 6} \right)\left( 3 \right) + \left( 2 \right)\left( 4 \right) + 11}}{{\sqrt {{3^2} + {{\left( { - 6} \right)}^2} + {2^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{6 - 18 + 8 + 11}}{{\sqrt {9 + 36 + 4} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{7}{{\sqrt {49} }}} \right|\]
\[ \Rightarrow D = \dfrac{7}{7}\]
\[ \Rightarrow D = 1\]
Thus, the distance between the point \[\left( {2,3,4} \right)\] and the plane \[3x - 6y + 2x + 11 = 0\] is 1 unit.
Option ‘A’ is correct
Note: Sometimes students use the formula of the distance between the plane and a point as \[D = \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. If the value \[a{x_1} + b{y_1} + c{z_1} + d < 0\], then they will get the \[D < 0\]. This is wrong because the distance is always positive. So, remember to apply the modulus function on the right-hand side.
Formula Used:The smallest distance between a point \[\left( {{x_1},{y_1},{z_1}} \right)\] and a plane \[ax + by + cz = d\] 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:The given point is \[\left( {2,3,4} \right)\] and the equation of the plane is \[3x - 6y + 2x + 11 = 0\].
By comparing the given information with the formula of the distance between the plane and a point, we get
\[{x_1} = 2,{y_1} = 3,{z_1} = 4\] and \[a = 3,b = - 6,c = 2\].
Now substitute the values in the formula of the distance between the plane and a point \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
\[D = \left| {\dfrac{{\left( 3 \right)\left( 2 \right) + \left( { - 6} \right)\left( 3 \right) + \left( 2 \right)\left( 4 \right) + 11}}{{\sqrt {{3^2} + {{\left( { - 6} \right)}^2} + {2^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{6 - 18 + 8 + 11}}{{\sqrt {9 + 36 + 4} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{7}{{\sqrt {49} }}} \right|\]
\[ \Rightarrow D = \dfrac{7}{7}\]
\[ \Rightarrow D = 1\]
Thus, the distance between the point \[\left( {2,3,4} \right)\] and the plane \[3x - 6y + 2x + 11 = 0\] is 1 unit.
Option ‘A’ is correct
Note: Sometimes students use the formula of the distance between the plane and a point as \[D = \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. If the value \[a{x_1} + b{y_1} + c{z_1} + d < 0\], then they will get the \[D < 0\]. This is wrong because the distance is always positive. So, remember to apply the modulus function on the right-hand side.
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