Question
Question: The distance between the planes \(x + 2y + 3z + 7 = 0\) and \(2x + 4y + 6z + 7 = 0\) is: A. \(\dfr...
The distance between the planes x+2y+3z+7=0 and 2x+4y+6z+7=0 is:
A. 227
B. 27
C. 27
D. 227
Solution
In the given question first we will divide the second equation of plane by2. By doing this first three terms of both equations will be equal. After this we will compare these equation from ax+by+cz+D1=0andax+by+cz+D2=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 a2+b2+c2∣D1−D2∣
Complete step by step Answer:
From the question:
Given that:
Two equations of plane are:
x+2y+3z+7=0..........(1)
2x+4y+6z+7=0...........(2)
From the given equations, equation of second plane is by dividing equation (2) by2 and can be written as:
x+2y+3z+27=0.........(3)
This can be done by taking 2 as common from equation (2)
Now the first three terms of equation (1)&(3) are equal
Now we can find the distance:
The distance between parallel planes is:
ax+by+cz+D1=0
And
ax+by+cz+D2=0
Where
a=1 b=3 c=3 D1=7 D2=27
From equation one and equation three.
Now the formula of distance between two planes is:
a2+b2+c2∣D1−D2∣
Put all the values in above formula:
We get:
Hence the correct answer is option A.
Note: First of all we have to divide the second equation of plane by2, without this we cannot find the value of distance. We have to remember that to compare both equation from ax+by+cz+D1=0 and ax+by+cz+D2=0, also learn and remember the formula of distance between two planes i.e. a2+b2+c2∣D1−D2∣. Thus we get the correct answer.