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Question Number 110320 by bemath last updated on 28/Aug/20

find the point of intersection of the line r^→ =(1−2t,3+4t,t) and the plane 3x−2y+5z=15

$${find}\:{the}\:{point}\:{of}\:{intersection} \\ $$$${of}\:{the}\:{line}\:\overset{\rightarrow} {{r}}=\left(\mathrm{1}−\mathrm{2}{t},\mathrm{3}+\mathrm{4}{t},{t}\right) \\ $$$${and}\:{the}\:{plane}\:\mathrm{3}{x}−\mathrm{2}{y}+\mathrm{5}{z}=\mathrm{15}\: \\ $$

Answered by mr W last updated on 28/Aug/20

3(1−2t)−2(3+4t)+5(t)=15 −9t=18 ⇒t=−2 1−2t=5 3+4t=−5 intersection point is (5,−5,−2)

$$\mathrm{3}\left(\mathrm{1}−\mathrm{2}{t}\right)−\mathrm{2}\left(\mathrm{3}+\mathrm{4}{t}\right)+\mathrm{5}\left({t}\right)=\mathrm{15} \\ $$$$−\mathrm{9}{t}=\mathrm{18} \\ $$$$\Rightarrow{t}=−\mathrm{2} \\ $$$$\mathrm{1}−\mathrm{2}{t}=\mathrm{5} \\ $$$$\mathrm{3}+\mathrm{4}{t}=−\mathrm{5} \\ $$$${intersection}\:{point}\:{is}\:\left(\mathrm{5},−\mathrm{5},−\mathrm{2}\right) \\ $$

Commented by I want to learn more last updated on 28/Aug/20

Sir please help me check the permutation question in Q110156 and 110157

$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}\:\mathrm{the}\:\mathrm{permutation}\:\mathrm{question}\:\mathrm{in}\:\:\mathrm{Q110156}\:\:\mathrm{and}\:\:\mathrm{110157} \\ $$

Answered by Rio Michael last updated on 28/Aug/20

x = 1−2t y = 3 + 4t z = t substituting ⇒ 3(1−2t)−2(3 + 4t) + 5t = 15 3−6t−6−8t + 5t = 15 ⇒ −9t = 18 hence t = −2 ⇒ point of intersection is (5,−5,−2)

$${x}\:=\:\mathrm{1}−\mathrm{2}{t} \\ $$$${y}\:=\:\mathrm{3}\:+\:\mathrm{4}{t} \\ $$$${z}\:=\:{t} \\ $$$$\mathrm{substituting}\:\Rightarrow\:\mathrm{3}\left(\mathrm{1}−\mathrm{2}{t}\right)−\mathrm{2}\left(\mathrm{3}\:+\:\mathrm{4}{t}\right)\:+\:\mathrm{5}{t}\:=\:\mathrm{15} \\ $$$$\mathrm{3}−\mathrm{6}{t}−\mathrm{6}−\mathrm{8}{t}\:+\:\mathrm{5}{t}\:=\:\mathrm{15} \\ $$$$\Rightarrow\:−\mathrm{9}{t}\:=\:\mathrm{18}\:\mathrm{hence}\:{t}\:=\:−\mathrm{2} \\ $$$$\Rightarrow\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{is}\:\:\left(\mathrm{5},−\mathrm{5},−\mathrm{2}\right) \\ $$

Answered by 1549442205PVT last updated on 28/Aug/20

The equation of the line is { ((x=1−2t)),((y=3+4t)),((z=t)) :} Replace into 3x−2y+5z=15 we have 3(1−2t)−2(3+4t)+5t=15 ⇔−9t=18⇒t=−2 Substituting into the system above we get x=5,y=−5,−2 Therefore,the intersection point of the line and the plane is the point M(5,−5,−2)

$$\mathrm{The}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{is}\:\begin{cases}{\mathrm{x}=\mathrm{1}−\mathrm{2t}}\\{\mathrm{y}=\mathrm{3}+\mathrm{4t}}\\{\mathrm{z}=\mathrm{t}}\end{cases} \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\mathrm{3}{x}−\mathrm{2}{y}+\mathrm{5}{z}=\mathrm{15}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{3}\left(\mathrm{1}−\mathrm{2t}\right)−\mathrm{2}\left(\mathrm{3}+\mathrm{4t}\right)+\mathrm{5t}=\mathrm{15} \\ $$$$\Leftrightarrow−\mathrm{9t}=\mathrm{18}\Rightarrow\mathrm{t}=−\mathrm{2} \\ $$$$\mathrm{Substituting}\:\mathrm{into}\:\mathrm{the}\:\mathrm{system}\:\mathrm{above} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{x}=\mathrm{5},\mathrm{y}=−\mathrm{5},−\mathrm{2} \\ $$$$\mathrm{Therefore},\mathrm{the}\:\mathrm{intersection}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{line}\:\mathrm{and}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{M}\left(\mathrm{5},−\mathrm{5},−\mathrm{2}\right) \\ $$$$ \\ $$

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