Question
Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.
✓
Answer
The given equation are
2x − y + 3z = 4 ....(1)
The second equation of the plane is
2x − y + 3z = 18 .....(2)
We know that distance between two planes ax + by + cz = d1 and ax + by + cz = d2 is $\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance is
$\frac{|18-4|}{\sqrt{2^2+(-1)^2+3^2}}$
$=\frac{|14|}{\sqrt{4+1+9}}$
$=\frac{14}{\sqrt{14}}$
$=\sqrt{14}\text{ units}$
2x − y + 3z = 4 ....(1)
The second equation of the plane is
2x − y + 3z = 18 .....(2)
We know that distance between two planes ax + by + cz = d1 and ax + by + cz = d2 is $\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance is
$\frac{|18-4|}{\sqrt{2^2+(-1)^2+3^2}}$
$=\frac{|14|}{\sqrt{4+1+9}}$
$=\frac{14}{\sqrt{14}}$
$=\sqrt{14}\text{ units}$
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