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# a))  show that the distance between parallel planes Ax+By+Cz+D1=0 and Ax+By+Cz+D2=0 in 3space is:d=l D1 -D2 l/$\sqrt{\ } \!\,$A^2+B^2+C^2b)) Use this result to find the distance between the parallel planes 3x-9y-12z+5=0  and x-3y-4z+6=0

Latika Leekha
6 years ago
Hello student,
First of all, select two arbitrary points P1(x1, y1, z1) and P2(x2, y2, z2) on
Ax + By + Cz + D1 = 0 and Ax + By + Cz + D2 = 0 respectively.
Now, therefore, the vector t = <x2-x1, y2-y1, z2-z1> is the vector starting at P1 and ending at P2.
Moreover, the projection of vector t onto the normal vector n = <a, b, c> is the distance between the two planes.
This length will therefore be given by the modulus of the projection of t onto n and is therefore given as under:

Now, since the points lie on the given planes so we have
Ax1 + By1 + Cz1 = -D1 and Ax2 + By2 + Cz2 = -D2.
Hence, using these the above expression reduces to

This is the required formula.