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Chapter 10: Q .42. (page 849)

Show that the planes given by y − 7z = 16 and 2y − 14z = 5 are parallel, and find the distance between the planes.

Short Answer

Expert verified

Therefore, the distance from the point to the plane is112613units.

Step by step solution

01

Step 1:Given information

The planes given by y − 7z = 16 and 2y − 14z = 5

02

Step 2:Calculation 

Consider the pointP(1,2,-3)and the line with the equationr(t)=⟨2+5t,-3+t,-4t⟩and the

plane with the equationx-3y+4z=5.

The objective is to find the distance from the pointP(1,2,-3)to the plane with the equation

x-3y+4z=5

The normal vector of the planex-3y+4z=5is:

N=⟨1,-3,4⟩

The pointR=(5,0,0)is on the planex-3y+4z=5.

The vectorRP→is given by:

RP→=⟨1-5,2-0,-3-0⟩

=⟨-4,2,-3⟩

The distance fromPto the plane is:

compNRP→=|N·RP|‖N‖

=|⟨1,-3,4⟩·⟨-4,2,-3⟩|‖⟨1,-3,4⟩‖

=|-4-6-12|1+9+16

=|-22|26

=2226

=112613

Therefore, the distance from the point to the plane is112613units.

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In Exercises 36–41 use the given sets of points to find:

(a) A nonzero vector N perpendicular to the plane determined by the points.

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(c) The area of the triangle determined by the points.

P(4,3,−2),Q(0,0,3),R(−1,3,6)

Fill in the blanks to complete each of the following theorem statements:

For ε>0,f(x)∈(L-ε,L+ε)if and only if<ε.
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