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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/ A^2+B^2+C^2 b)) Use this result to find the distance between the parallel planes 3x-9y-12z+5=0 and x-3y-4z+6=0

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^2


b)) Use this result to find the distance between the parallel planes 3x-9y-12z+5=0  and x-3y-4z+6=0

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1 Answers

Latika Leekha
askIITians Faculty 165 Points
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:
\left | \frac{t.n}{|n|} \right |= \frac{\left | A(x_{2}-x_{1})+B(y_{2}-y_{1})+C(z_{2}-z_{1})\right |}{\sqrt{A^{2}+B^{2}+C^{2}}}
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
\frac{\left | -D_{2}+D_{1}\right |}{\sqrt{A^{2}+B^{2}+C^{2}}}
This is the required formula.

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