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Calculus IIIChapter 1: 3-Dimensional SpaceSection 1-1

The 3-D Coordinate System — Question 6

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Question 6

Two points determine a directed line segment: A=(−4,1,7),B=(6,−3,−1).A=(-4,1,7),\qquad B=(6,-3,-1).

Tasks

  1. Find the point PP that divides AB¯\overline{AB} internally in the ratio AP:PB=2:3.AP:PB=2:3.

  2. Find the point QQ that divides the same line externally in the ratio AQ:QB=2:3.AQ:QB=2:3.

  3. Verify both ratios using three-dimensional distances, and distinguish clearly between internal and external division.

Original worksheet page 1: question and worked solution for 1-1-006
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Question 6 – Solution

Strategy Parameterize the line by X(t)=A+t(B−A)X(t)=A+t(B-A). Internal division has 0<t<10<t<1; external division requires tt outside that interval.

See the diagram in the original worksheet below.

Internal point Since B−A=(10,−4,−8)B-A=(10,-4,-8) and AP/AB=2/(2+3)=2/5AP/AB=2/(2+3)=2/5, P=A+25(B−A)=(−4,1,7)+(4,−8/5,−16/5)=(0,−3/5,19/5).P=A+\frac 25(B-A)=(-4,1,7)+(4,-8/5,-16/5)=\boxed{(0,-3/5,19/5)}.

External point For Q=A+t(B−A)Q=A+t(B-A) beyond AA, take t<0t<0. The ratio is AQQB=−t1−t=23,\frac{AQ}{QB}=\frac{-t}{1-t}=\frac 23, which gives t=−2t=-2. Thus Q=A−2(B−A)=(−24,9,23).Q=A-2(B-A)=\boxed{(-24,9,23)}.

Verification Because P−A=(2/5)(B−A)P-A=(2/5)(B-A) and B−P=(3/5)(B−A)B-P=(3/5)(B-A), their lengths have ratio 2:32:3. Also Q−A=−2(B−A)Q-A=-2(B-A) and B−Q=3(B−A)B-Q=3(B-A), again giving AQ:QB=2:3AQ:QB=2:3.

Original worksheet page 2: question and worked solution for 1-1-006

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