Question

Find the length of the perpendicular from the origin to the plane \[3x + 4y + 12z = 52\].
A. 3
B. \[ - 4\]
C. 5
D. None of these

Answer 1

Hint: Here, an equation of the plane is given. First, calculate the direction cosines of the given plane. Then, rewrite the equation in the form of direction cosines. The value of the constant term on the right-hand side is the required distance.

Formula used: The direction cosines of the plane \[ax + by + cz + d = 0\] are: \[\left( {\dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right)\]


Complete step by step solution: Given:
The plane \[3x + 4y + 12z = 52\] is perpendicular to the origin.

From the equation of the plane, we get
The direction ratios of the plane are \[\left( {3,4,12} \right)\].
Let’s calculate the direction cosines of the plane.
Apply the formula for the direction cosines of the plane.
We get,
\[\left( {\dfrac{3}{{\sqrt {{3^2} + {4^2} + {{12}^2}} }},\dfrac{4}{{\sqrt {{3^2} + {4^2} + {{12}^2}} }},\dfrac{{12}}{{\sqrt {{3^2} + {4^2} + {{12}^2}} }}} \right)\]
\[ \Rightarrow \left( {\dfrac{3}{{\sqrt {9 + 16 + 144} }},\dfrac{4}{{\sqrt {9 + 16 + 144} }},\dfrac{{12}}{{\sqrt {9 + 16 + 144} }}} \right)\]
\[ \Rightarrow \left( {\dfrac{3}{{\sqrt {169} }},\dfrac{4}{{\sqrt {169} }},\dfrac{{12}}{{\sqrt {169} }}} \right)\]
\[ \Rightarrow \left( {\dfrac{3}{{13}},\dfrac{4}{{13}},\dfrac{{12}}{{13}}} \right)\]

Now divide the equation of the plane by \[13\].
 \[\dfrac{3}{{13}}x + \dfrac{4}{{13}}y + \dfrac{{12}}{{13}}z = \dfrac{{52}}{{13}}\]
\[ \Rightarrow \dfrac{3}{{13}}x + \dfrac{4}{{13}}y + \dfrac{{12}}{{13}}z = 4\]
Thus, the length of the perpendicular from the origin to the plane \[3x + 4y + 12z = 52\] is 4 units.


Thus, Option (D) is correct.
Note: The other way to calculate the length of the perpendicular from the origin to the plane \[ax + by + cz + d = 0\] use the following formula.
The length of the perpendicular from the origin to the plane: \[l = \left| {\dfrac{{ - d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]