Question-46095

Question Number 46095 by Meritguide1234 last updated on 21/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

The eqn of stline is ((x−2)/1)=((y−1)/2)=((z−2)/3)=k eqn plane x+y+z=8 (2,1,2) is a point on stline (k+2,2k+1,3k+2) is the point where st line intersect plane. so k+2+2k+1+3k+2=8 6k+5=8 k=(1/2) ((5/2),2,(7/2)) is the point where st line intersect plane now we are going to find image of(2,1,2) wrt plane eqn st line passing through (2,1,2) and parallel to the normal on the plane is ((x−2)/1)=((y−1)/1)=((z−2)/1)=α (α+2,α+1,α+2) lies on plane α+2+α+1+α+2=8 3α=3 α=1 (3,2,3) (2,1,2) (3,2,3) (a,b,c) here[(3,2,3) is foot of perpendicular and (a,b,c) is image of (2,1,2) (3,2,3) is mid point of (2,1,2) and (a,b,c) so ((2+a)/2)=3 ((1+b)/2)=2 ((2+c)/2)=3 a=4 b=3 c=4 SO The eqn of image of st line is the st line passing through ((5/2),2,(7/2)) and(4,3,4) ((x−(5/2))/(4−(5/2)))=((y−2)/(3−2))=((z−(7/2))/(4−(7/2))) ((x−2.5)/(1.5))=((y−2)/1)=((z−3.5)/(0.5)) or ((x−4)/(1.5))=((y−3)/1)=((z−4)/(0.5)) ←image line PROJECTION line passing through ((5/2),(2/1),(7/2)) and foot of perpendicular(3,2,3) ((x−3)/(3−2.5))=((y−2)/(2−2))=((z−3)/(3−3.5)) ((x−3)/(0.5))=((y−2)/0)=((z−3)/(−0.5)) ← pls check...

$${The}\:{eqn}\:{of}\:{stline}\:{is} \\ $$$$\frac{{x}−\mathrm{2}}{\mathrm{1}}=\frac{{y}−\mathrm{1}}{\mathrm{2}}=\frac{{z}−\mathrm{2}}{\mathrm{3}}={k} \\ $$$${eqn}\:{plane}\:{x}+{y}+{z}=\mathrm{8} \\ $$$$\left(\mathrm{2},\mathrm{1},\mathrm{2}\right)\:{is}\:{a}\:{point}\:{on}\:{stline} \\ $$$$\left({k}+\mathrm{2},\mathrm{2}{k}+\mathrm{1},\mathrm{3}{k}+\mathrm{2}\right)\:{is}\:{the}\:{point}\:{where}\:{st}\:{line} \\ $$$${intersect}\:{plane}. \\ $$$${so}\:{k}+\mathrm{2}+\mathrm{2}{k}+\mathrm{1}+\mathrm{3}{k}+\mathrm{2}=\mathrm{8} \\ $$$$\mathrm{6}{k}+\mathrm{5}=\mathrm{8}\:\:\:{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{2}},\mathrm{2},\frac{\mathrm{7}}{\mathrm{2}}\right)\:{is}\:{the}\:{point}\:{where}\:{st}\:{line}\:{intersect}\:{plane} \\ $$$${now}\:{we}\:{are}\:{going}\:{to}\:{find}\:{image}\:{of}\left(\mathrm{2},\mathrm{1},\mathrm{2}\right)\:{wrt}\:{plane} \\ $$$${eqn}\:{st}\:{line}\:{passing}\:{through}\:\left(\mathrm{2},\mathrm{1},\mathrm{2}\right)\:{and}\:{parallel} \\ $$$${to}\:{the}\:{normal}\:{on}\:{the}\:{plane}\:{is} \\ $$$$\frac{{x}−\mathrm{2}}{\mathrm{1}}=\frac{{y}−\mathrm{1}}{\mathrm{1}}=\frac{{z}−\mathrm{2}}{\mathrm{1}}=\alpha \\ $$$$\left(\alpha+\mathrm{2},\alpha+\mathrm{1},\alpha+\mathrm{2}\right)\:{lies}\:{on}\:{plane} \\ $$$$\alpha+\mathrm{2}+\alpha+\mathrm{1}+\alpha+\mathrm{2}=\mathrm{8}\:\:\mathrm{3}\alpha=\mathrm{3}\:\:\:\alpha=\mathrm{1} \\ $$$$\left(\mathrm{3},\mathrm{2},\mathrm{3}\right) \\ $$$$\left(\mathrm{2},\mathrm{1},\mathrm{2}\right)\:\:\:\left(\mathrm{3},\mathrm{2},\mathrm{3}\right)\:\:\:\left({a},{b},{c}\right) \\ $$$${here}\left[\left(\mathrm{3},\mathrm{2},\mathrm{3}\right)\:{is}\:{foot}\:{of}\:{perpendicular}\:{and}\right. \\ $$$$\left({a},{b},{c}\right)\:\:{is}\:{image}\:{of}\:\left(\mathrm{2},\mathrm{1},\mathrm{2}\right) \\ $$$$\left(\mathrm{3},\mathrm{2},\mathrm{3}\right)\:{is}\:{mid}\:{point}\:{of}\:\left(\mathrm{2},\mathrm{1},\mathrm{2}\right)\:{and}\:\left({a},{b},{c}\right) \\ $$$${so}\:\frac{\mathrm{2}+{a}}{\mathrm{2}}=\mathrm{3}\:\:\:\:\:\:\frac{\mathrm{1}+{b}}{\mathrm{2}}=\mathrm{2}\:\:\:\:\frac{\mathrm{2}+{c}}{\mathrm{2}}=\mathrm{3} \\ $$$${a}=\mathrm{4}\:\:\:\:\:\:\:{b}=\mathrm{3}\:\:\:\:\:{c}=\mathrm{4} \\ $$$$\boldsymbol{{S}}{O}\:{The}\:{eqn}\:{of}\:{image}\:{of}\:{st}\:{line}\:{is}\:{the}\:{st}\:{line} \\ $$$${passing}\:{through}\:\left(\frac{\mathrm{5}}{\mathrm{2}},\mathrm{2},\frac{\mathrm{7}}{\mathrm{2}}\right)\:{and}\left(\mathrm{4},\mathrm{3},\mathrm{4}\right) \\ $$$$\frac{{x}−\frac{\mathrm{5}}{\mathrm{2}}}{\mathrm{4}−\frac{\mathrm{5}}{\mathrm{2}}}=\frac{{y}−\mathrm{2}}{\mathrm{3}−\mathrm{2}}=\frac{{z}−\frac{\mathrm{7}}{\mathrm{2}}}{\mathrm{4}−\frac{\mathrm{7}}{\mathrm{2}}} \\ $$$$\frac{{x}−\mathrm{2}.\mathrm{5}}{\mathrm{1}.\mathrm{5}}=\frac{{y}−\mathrm{2}}{\mathrm{1}}=\frac{{z}−\mathrm{3}.\mathrm{5}}{\mathrm{0}.\mathrm{5}} \\ $$$${or}\:\frac{{x}−\mathrm{4}}{\mathrm{1}.\mathrm{5}}=\frac{{y}−\mathrm{3}}{\mathrm{1}}=\frac{{z}−\mathrm{4}}{\mathrm{0}.\mathrm{5}}\:\leftarrow{image}\:{line} \\ $$$$\boldsymbol{{P}}{ROJECTION}\:{line}\:{passing}\:{through} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{1}},\frac{\mathrm{7}}{\mathrm{2}}\right)\:{and}\:{foot}\:{of}\:{perpendicular}\left(\mathrm{3},\mathrm{2},\mathrm{3}\right) \\ $$$$\frac{{x}−\mathrm{3}}{\mathrm{3}−\mathrm{2}.\mathrm{5}}=\frac{{y}−\mathrm{2}}{\mathrm{2}−\mathrm{2}}=\frac{{z}−\mathrm{3}}{\mathrm{3}−\mathrm{3}.\mathrm{5}} \\ $$$$\frac{{x}−\mathrm{3}}{\mathrm{0}.\mathrm{5}}=\frac{{y}−\mathrm{2}}{\mathrm{0}}=\frac{{z}−\mathrm{3}}{−\mathrm{0}.\mathrm{5}}\:\leftarrow\:{pls}\:{check}… \\ $$$$ \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18