Question

Find the distance of the plane \[6x - 3y + 2z - 14 = 0\]from its origin.

a) 2
b) 1
c) 14
d) 8

Answer 1

Hint: We can use the formula for finding the perpendicular distance between the origin and the plane to find the required distance.



Formula Used:The perpendicular distance from the origin to the plane \[ax + by + cz + d = 0\]is given by \[\dfrac{{\left| d \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]



Complete step by step solution:The equation of the given plane is \[6x - 3y + 2z - 14 = 0.\]

The perpendicular distance from the origin to the plane \[6x - 3y + 2z - 14 = 0\] \[ = \dfrac{{\left| { - 14} \right|}}{{\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }}\]

      \[ = \dfrac{{14}}{{\sqrt {36 + 9 + 4} }}\]

      \[ = \dfrac{{14}}{{\sqrt {49} }}\]

      \[ = \dfrac{{14}}{7}\]

       = 2



Option ‘a’ is correct



Note:This question can also be solved using the following method.

The equation of the given plane is \[6x - 3y + 2z - 14 = 0.\]

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


The perpendicular distance from the point\[(0,0,0)\]to the plane \[6x - 3y + 2z - 14 = 0\] \[ = \left| {\dfrac{{6\left( 0 \right) - 3\left( 0 \right) + 2\left( 0 \right) - 14}}{{\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }}} \right|\]

                    \[ = \left| {\dfrac{{ - 14}}{{\sqrt {36 + 9 + 4} }}} \right|\]

\[ = \left| {\dfrac{{ - 14}}{{\sqrt {49} }}} \right|\]

\[ = \dfrac{{14}}{7}\]

= 2

So, the correct choice is a.