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Question Number 75317 by vishalbhardwaj last updated on 10/Dec/19

Find the image of the line ((x−2)/(−1)) = ((y−3)/2) = ((z−4)/3) in the plane 2x+3y−4z +3=0.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\: \\ $$$$\frac{{x}−\mathrm{2}}{−\mathrm{1}}\:=\:\frac{\mathrm{y}−\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{z}−\mathrm{4}}{\mathrm{3}}\:\:\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane}\: \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}−\mathrm{4}{z}\:+\mathrm{3}=\mathrm{0}. \\ $$

Commented by vishalbhardwaj last updated on 10/Dec/19

yes sir, I corrected it

$$\mathrm{yes}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\ $$

Answered by vishalbhardwaj last updated on 10/Dec/19

please solve this

$$\mathrm{please}\:\mathrm{solve}\:\mathrm{this} \\ $$

Answered by mr W last updated on 10/Dec/19

line ((x−2)/(−1)) = ((y−3)/2) = ((z−4)/3)=λ ⇒x=−λ+2 ⇒y=2λ+3 ⇒z=3λ+4 using λ=1 we get a point P on the line P(1,5,7). say the image of point P in the plane is P′(u,v,w) (x,y,z)=(1,5,7)+k(2,3,−4) 2(1+2k)+3(5+3k)−4(7−4k)+3=0 ⇒k=(8/3) (u,v,w)=(1,5,7)+((16)/3)(2,3,−4)=(((35)/3),21,((−43)/3)) intersection point Q of line and plane: 2(−λ+2)+3(2λ+3)−4(3λ+4)+3=0 ⇒λ=0 ⇒Q(2,3,4) the image of line in plane is the line P′Q: ((x−2)/(((35)/3)−2))=((y−3)/(21−3))=((z−4)/(−((43)/3)−4)) ⇒((x−2)/((29)/3))=((y−3)/(18))=((z−4)/(−((55)/3)))

$${line} \\ $$$$\frac{{x}−\mathrm{2}}{−\mathrm{1}}\:=\:\frac{\mathrm{y}−\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{z}−\mathrm{4}}{\mathrm{3}}=\lambda \\ $$$$\Rightarrow{x}=−\lambda+\mathrm{2} \\ $$$$\Rightarrow{y}=\mathrm{2}\lambda+\mathrm{3} \\ $$$$\Rightarrow{z}=\mathrm{3}\lambda+\mathrm{4} \\ $$$${using}\:\lambda=\mathrm{1}\:{we}\:{get}\:{a}\:{point}\:{P}\:{on}\:{the} \\ $$$${line}\:{P}\left(\mathrm{1},\mathrm{5},\mathrm{7}\right). \\ $$$${say}\:{the}\:{image}\:{of}\:{point}\:{P}\:{in}\:{the} \\ $$$${plane}\:{is}\:{P}'\left({u},{v},{w}\right) \\ $$$$\left({x},{y},{z}\right)=\left(\mathrm{1},\mathrm{5},\mathrm{7}\right)+{k}\left(\mathrm{2},\mathrm{3},−\mathrm{4}\right) \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{2}{k}\right)+\mathrm{3}\left(\mathrm{5}+\mathrm{3}{k}\right)−\mathrm{4}\left(\mathrm{7}−\mathrm{4}{k}\right)+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left({u},{v},{w}\right)=\left(\mathrm{1},\mathrm{5},\mathrm{7}\right)+\frac{\mathrm{16}}{\mathrm{3}}\left(\mathrm{2},\mathrm{3},−\mathrm{4}\right)=\left(\frac{\mathrm{35}}{\mathrm{3}},\mathrm{21},\frac{−\mathrm{43}}{\mathrm{3}}\right) \\ $$$$ \\ $$$${intersection}\:{point}\:{Q}\:{of}\:{line}\:{and}\:{plane}: \\ $$$$\mathrm{2}\left(−\lambda+\mathrm{2}\right)+\mathrm{3}\left(\mathrm{2}\lambda+\mathrm{3}\right)−\mathrm{4}\left(\mathrm{3}\lambda+\mathrm{4}\right)+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\lambda=\mathrm{0} \\ $$$$\Rightarrow{Q}\left(\mathrm{2},\mathrm{3},\mathrm{4}\right) \\ $$$${the}\:{image}\:{of}\:{line}\:{in}\:{plane}\:{is}\:{the} \\ $$$${line}\:{P}'{Q}: \\ $$$$\frac{{x}−\mathrm{2}}{\frac{\mathrm{35}}{\mathrm{3}}−\mathrm{2}}=\frac{{y}−\mathrm{3}}{\mathrm{21}−\mathrm{3}}=\frac{{z}−\mathrm{4}}{−\frac{\mathrm{43}}{\mathrm{3}}−\mathrm{4}} \\ $$$$\Rightarrow\frac{{x}−\mathrm{2}}{\frac{\mathrm{29}}{\mathrm{3}}}=\frac{{y}−\mathrm{3}}{\mathrm{18}}=\frac{{z}−\mathrm{4}}{−\frac{\mathrm{55}}{\mathrm{3}}} \\ $$

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