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Chapter 10: Problem 51

Find the distance between the given objects. The planes \(x+3 y-2 z=3\) and \(x+3 y-2 z=1\)

Short Answer

Expert verified
The distance between the given planes is \(\frac{2}{\sqrt{14}}.\)

Step by step solution

01

Identify the format of the planes

Identify that the given planes are in the format 'ax+by+cz=d'. In the given planes, \(a=1, b=3, c=-2\) for both planes, which means that they are parallel.
02

Subtract the constant terms from the away plane from the near plane

Next, subtract the constant terms of each plane from each other which refer to the 'd' in 'ax+by+cz=d'. For the given planes, the constants are 3 and 1. So, subtract 1 from 3, which gives 2.
03

Calculate the distance

The distance between two parallel planes is given by \(d=|d1-d2|/sqrt(a^2+b^2+c^2)\). So, substitute \(a=1, b=3, c=-2, d1=3, d2=1\), the distance is \(|3-1|/sqrt(1^2+3^2+(-2)^2) = 2/sqrt(14)\)

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Most popular questions from this chapter

Find the intersection of the planes. $$3 x+y-z=2 \text { and } 2 x-3 y+z=-1$$

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Find an equation of the given plane. The plane containing the points (-2,2,0),(-2,3,2) and (1,2,2)

Sketch the given traces on a single three-dimensional coordinate system. $$z=x^{2}+y^{2} ; y=0, y=1, y=2$$
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