Question Number 55475 by Knight last updated on 25/Feb/19
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{passing}\:\mathrm{through}\:\mathrm{the}\:\mathrm{points} \\ $$$$\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)\:\mathrm{and}\:\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)\:\mathrm{and}\:\mathrm{makes} \\ $$$$\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\:\frac{\pi}{\mathrm{4}}\:\mathrm{with}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{x}+\mathrm{y}\:=\:\mathrm{3} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Feb/19
![let eqn of plane a(x−1)+b(y−0)+c(z−0)=0 which passes through (0,1,0) a(0−1)+b(1−0)+c(0−0)=0 −a+b=0 [a=b] angle between a(x−1)+b(y−0)+c(z−0)=0 and x+y+0.z=3 is cos(π/4)=((a×1+b×1+c×0)/( (√(a^2 +b^2 +c^2 )) ×(√(1^2 +1^2 +0^2 )) )) nos (1/( (√2)))=((a+b)/( (√(a^2 +b^2 +c^2 )) ×(√2))) 1=((a+a)/( (√(a^2 +a^2 +c^2 )))) 4a^2 =2a^2 +c^2 c=±a(√2) so the eqn of planes are a(x−1)+b(y−0)+c(z−0)=0 a(x−1)+a(y−0)+a(√2) (z−0)=0 x−1+y+(√2) z=0 x+y+(√2) z=1 and a(x−1)+b(y−0)−a(√2) (z−0)=0 x−1+y−(√2) z=0 x+y−(√2) z=1](https://www.tinkutara.com/question/Q55477.png)
$${let}\:{eqn}\:{of}\:{plane}\:{a}\left({x}−\mathrm{1}\right)+{b}\left({y}−\mathrm{0}\right)+{c}\left({z}−\mathrm{0}\right)=\mathrm{0} \\ $$$${which}\:{passes}\:{through}\:\left(\mathrm{0},\mathrm{1},\mathrm{0}\right) \\ $$$${a}\left(\mathrm{0}−\mathrm{1}\right)+{b}\left(\mathrm{1}−\mathrm{0}\right)+{c}\left(\mathrm{0}−\mathrm{0}\right)=\mathrm{0} \\ $$$$−{a}+{b}=\mathrm{0}\:\:\:\:\:\left[{a}={b}\right] \\ $$$${angle}\:{between}\:{a}\left({x}−\mathrm{1}\right)+{b}\left({y}−\mathrm{0}\right)+{c}\left({z}−\mathrm{0}\right)=\mathrm{0} \\ $$$${and}\:{x}+{y}+\mathrm{0}.{z}=\mathrm{3}\:{is} \\ $$$${cos}\frac{\pi}{\mathrm{4}}=\frac{{a}×\mathrm{1}+{b}×\mathrm{1}+{c}×\mathrm{0}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\:×\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} }\:} \\ $$$${nos} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{{a}+{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\:×\sqrt{\mathrm{2}}} \\ $$$$\mathrm{1}=\frac{{a}+{a}}{\:\sqrt{{a}^{\mathrm{2}} +{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${c}=\pm{a}\sqrt{\mathrm{2}}\: \\ $$$${so}\:{the}\:{eqn}\:{of}\:{planes}\:{are} \\ $$$${a}\left({x}−\mathrm{1}\right)+{b}\left({y}−\mathrm{0}\right)+{c}\left({z}−\mathrm{0}\right)=\mathrm{0} \\ $$$${a}\left({x}−\mathrm{1}\right)+{a}\left({y}−\mathrm{0}\right)+{a}\sqrt{\mathrm{2}}\:\left({z}−\mathrm{0}\right)=\mathrm{0} \\ $$$${x}−\mathrm{1}+{y}+\sqrt{\mathrm{2}}\:{z}=\mathrm{0} \\ $$$${x}+{y}+\sqrt{\mathrm{2}}\:{z}=\mathrm{1} \\ $$$${and} \\ $$$${a}\left({x}−\mathrm{1}\right)+{b}\left({y}−\mathrm{0}\right)−{a}\sqrt{\mathrm{2}}\:\left({z}−\mathrm{0}\right)=\mathrm{0} \\ $$$${x}−\mathrm{1}+{y}−\sqrt{\mathrm{2}}\:{z}=\mathrm{0} \\ $$$${x}+{y}−\sqrt{\mathrm{2}}\:{z}=\mathrm{1} \\ $$$$ \\ $$