Question

If a plane cuts off intercepts \[ - 6,{\rm{ }}3,{\rm{ }}4\] from the co-ordinate axes, then find the length of the perpendicular from the origin to the plane.
A. \[\dfrac{1}{{\sqrt {61} }}\]
B. \[\dfrac{{13}}{{\sqrt {61} }}\]
C. \[\dfrac{{12}}{{\sqrt {29} }}\]
D. \[\dfrac{5}{{\sqrt {41} }}\]

Answer 1

Hint: First, consider the equation of plane as \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]. Then, substitute the values of the intercepts in the equation of plane and simplify it. After that, calculate the distance from the origin to the plane by using the formula of the smallest distance between a point and plane and get the required answer.



Formula Used:The equation of plane: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
The smallest distance between a point \[\left( {p,q,r} \right)\] and a plane \[ax + by + cz = d\] is: \[D = \left| {\dfrac{{ap + bq + cr + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]



Complete step by step solution:Given:
The cut off intercepts from the coordinate axes are: \[ - 6,{\rm{ }}3,{\rm{ }}4\]

Let consider,
The general equation of a plane is \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\].
We have, x-intercept: \[a = - 6\], y-intercept: \[b = 3\] and z-intercept: \[c = 4\].
Substitute these values in the above equation.
We get,
\[\dfrac{x}{{ - 6}} + \dfrac{y}{3} + \dfrac{z}{4} = 1\]
\[ \Rightarrow \dfrac{{ - 2x + 4y + 3z}}{{12}} = 1\]
\[ \Rightarrow - 2x + 4y + 3z = 12\]

We have to calculate the perpendicular distance between the origin and plane.
So, use the formula of the smallest distance between a plane and a point.
Here, \[\left( {p,q,r} \right) = \left( {0,0,0} \right)\] and the equation of plane is \[ - 2x + 4y + 3z = 12\].
Substitute the values in the formula \[D = \left| {\dfrac{{ap + bq + cr + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
We get,
\[D = \left| {\dfrac{{\left( { - 2} \right)\left( 0 \right) + \left( 4 \right)\left( 0 \right) + \left( 3 \right)\left( 0 \right) + 12}}{{\sqrt {{{\left( { - 2} \right)}^2} + {4^2} + {3^2}} }}} \right|\]
\[ \Rightarrow D = \dfrac{{12}}{{\sqrt {4 + 16 + 9} }}\]
\[ \Rightarrow D = \dfrac{{12}}{{\sqrt {29} }}\]
Thus, the perpendicular distance between the origin and the plane \[ - 2x + 4y + 3z = 12\] is \[\dfrac{{12}}{{\sqrt {29} }}\] units.



Option ‘C’ is correct


Note: The shortest distance between a point and any geometric figure is the perpendicular distance between the point and that geometric figure.