KCET · Maths · Three Dimensional Geometry
The distance between the two planes \(2 x+3 y+4 z=4\) and \(4 x+6 y+8 z=12\) is
- A 2 units
- B 8 units
- C \(\frac{2}{\sqrt{29}}\) unit
- D 4 units
Answer & Solution
Correct Answer
(C) \(\frac{2}{\sqrt{29}}\) unit
Step-by-step Solution
Detailed explanation
The equations of planes are
\(2 x+3 y+4 z=4\) ....(i)
\(4 x+6 y+8 z=12\)
\(2 x+3 y+4 z=6\) ....(ii)
It can be seen that the given planes are parallel. It is known that the distance between two parallel
planes is \(\left|\frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}\right|\).
So, \(\quad D=\left|\frac{6-4}{\sqrt{(2)^2+(3)^2+(4)^2}}\right|=\frac{2}{\sqrt{29}}\) units
\(2 x+3 y+4 z=4\) ....(i)
\(4 x+6 y+8 z=12\)
\(2 x+3 y+4 z=6\) ....(ii)
It can be seen that the given planes are parallel. It is known that the distance between two parallel
planes is \(\left|\frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}\right|\).
So, \(\quad D=\left|\frac{6-4}{\sqrt{(2)^2+(3)^2+(4)^2}}\right|=\frac{2}{\sqrt{29}}\) units
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