
The distance between the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ is:
A. $\dfrac{{\sqrt 7 }}{{2\sqrt 2 }}$
B. $\dfrac{7}{2}$
C. $\dfrac{{\sqrt 7 }}{2}$
D. $\dfrac{7}{{2\sqrt 2 }}$
Answer
511.8k+ views
Hint: In the given question first we will divide the second equation of plane by$2$. By doing this first three terms of both equations will be equal. After this we will compare these equation from $ax + by + cz + {D_1} = 0$and$ax + by + cz + {D_2} = 0$. After that we will apply the formula of distance between two planes, thus we will get the answer.
Formula Used: distance between two planes is \[\dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
Complete step by step Answer:
From the question:
Given that:
Two equations of plane are:
$x + 2y + 3z + 7 = 0..........\left( 1 \right)$
$2x + 4y + 6z + 7 = 0...........\left( 2 \right)$
From the given equations, equation of second plane is by dividing equation $\left( 2 \right)$ by$2$ and can be written as:
$x + 2y + 3z + \dfrac{7}{2} = 0.........\left( 3 \right)$
This can be done by taking $2$ as common from equation $\left( 2 \right)$
Now the first three terms of equation \[\left( 1 \right)\& \left( 3 \right)\] are equal
Now we can find the distance:
The distance between parallel planes is:
$ax + by + cz + {D_1} = 0$
And
$ax + by + cz + {D_2} = 0$
Where
$
a = 1 \\
b = 3 \\
c = 3 \\
{D_1} = 7 \\
{D_2} = \dfrac{7}{2} \\
$
From equation one and equation three.
Now the formula of distance between two planes is:
\[\dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
Put all the values in above formula:
We get:
\[
\dfrac{{\left| {7 - \dfrac{7}{2}} \right|}}{{\sqrt {{1^2} + {2^2} + {3^2}} }} \\
\Rightarrow \dfrac{{\dfrac{{\left| {14 - 7} \right|}}{2}}}{{\sqrt {1 + 4 + 9} }} \\
\Rightarrow \dfrac{{\sqrt 7 }}{{2\sqrt 2 }} \\
\]
Hence the correct answer is option A.
Note: First of all we have to divide the second equation of plane by$2$, without this we cannot find the value of distance. We have to remember that to compare both equation from $ax + by + cz + {D_1} = 0$ and $ax + by + cz + {D_2} = 0$, also learn and remember the formula of distance between two planes i.e. \[\dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. Thus we get the correct answer.
Formula Used: distance between two planes is \[\dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
Complete step by step Answer:
From the question:
Given that:
Two equations of plane are:
$x + 2y + 3z + 7 = 0..........\left( 1 \right)$
$2x + 4y + 6z + 7 = 0...........\left( 2 \right)$
From the given equations, equation of second plane is by dividing equation $\left( 2 \right)$ by$2$ and can be written as:
$x + 2y + 3z + \dfrac{7}{2} = 0.........\left( 3 \right)$
This can be done by taking $2$ as common from equation $\left( 2 \right)$
Now the first three terms of equation \[\left( 1 \right)\& \left( 3 \right)\] are equal
Now we can find the distance:
The distance between parallel planes is:
$ax + by + cz + {D_1} = 0$
And
$ax + by + cz + {D_2} = 0$
Where
$
a = 1 \\
b = 3 \\
c = 3 \\
{D_1} = 7 \\
{D_2} = \dfrac{7}{2} \\
$
From equation one and equation three.
Now the formula of distance between two planes is:
\[\dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
Put all the values in above formula:
We get:
\[
\dfrac{{\left| {7 - \dfrac{7}{2}} \right|}}{{\sqrt {{1^2} + {2^2} + {3^2}} }} \\
\Rightarrow \dfrac{{\dfrac{{\left| {14 - 7} \right|}}{2}}}{{\sqrt {1 + 4 + 9} }} \\
\Rightarrow \dfrac{{\sqrt 7 }}{{2\sqrt 2 }} \\
\]
Hence the correct answer is option A.
Note: First of all we have to divide the second equation of plane by$2$, without this we cannot find the value of distance. We have to remember that to compare both equation from $ax + by + cz + {D_1} = 0$ and $ax + by + cz + {D_2} = 0$, also learn and remember the formula of distance between two planes i.e. \[\dfrac{{\left| {{D_1} - {D_2}} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. Thus we get the correct answer.
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