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Show that the planes given by 2x − 3y + 5z = 7 and −6x + 9y − 15z = 8 are parallel, and find the distance between the planes

Short Answer

Expert verified

Therefore, the distance from the pointP(1,2,-3)to the liner(t)=⟨2+5t,-3+t,-4t⟩is6.10units

Step by step solution

01

Step 1:Given information

the planes given by 2x − 3y + 5z = 7 and −6x + 9y − 15z = 8

02

Step 2:Calculation

Consider the pointP(1,2,-3)and the line with the equation r(t)=⟨2+5t,-3+t,-4t⟩and the plane with the equation x-3y+4z=5.

The pointP0(2,-3,0)corresponds tor(0)and the direction vector of the line isd=⟨5,1,-4⟩.

The norm of the direction vectord=⟨5,1,-4⟩is:

‖d‖=52+12+(-4)2

=25+1+16=42

The vectorP0P→is given by:

P0P→=⟨1-2,2-(-3),-3-0⟩

=⟨-1,5,-3⟩

The value ofd×P0P→is:

d×P→0P=ijk51-4-15-3

=-23i-19j+26k

=⟨-23,-19,26⟩

The distance from pointP(1,2,-3)to the liner(t)=⟨2+5t,-3+t,-4t⟩is:

d×P0P→‖d‖=‖⟨-23,-19,26⟩‖42

=156642

≈6.10units

Therefore, the distance from the pointP(1,2,-3)to the liner(t)=⟨2+5t,-3+t,-4t⟩is6.10units

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