A line contains the points P=(−6,−2,−3)P=(-6,-2,-3) and Q=(−4,−3,−6)Q=(-4,-3,-6). A vector equation of this line is →x=−−→OP+t−−→PQx→=OP→+tPQ→, where OO is the origin. If t=2t=2, what point does this give us from the equation?
A line contains the points P=(−6,−2,−3)P=(-6,-2,-3) and Q=(−4,−3,−6)Q=(-4,-3,-6). A vector equation of this line is →x=−−→OP+t−−→PQx→=OP→+tPQ→, where OO is the origin. If t=2t=2, what point does this give us from the equation?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A line contains the points P=(−6,−2,−3)P=(-6,-2,-3) and Q=(−4,−3,−6)Q=(-4,-3,-6). A
If t=2t=2, what point does this give us from the equation?
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