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Find the equation of the plane that contains the point A(2, 1, – 1) and is perpendicular to the line of intersection of the planes 2x + y – z = 3 and x + 2y + z = 2. Also find the angle between the plane thus obtained and the y-axis.

 

Answers (1)

Equation of plane that contains the point A(2,1,–1)

\\ a(x-2)+b(y-1)+c(z+1)=0

Accordinf to the condition;

\\ 2 a+b-c=0...(i) \\ a+2 b+c=0 ...(ii)\\

Equation(i) + Equation (ii)

3a + 3b = 0

a = -b 

c = -b =a 

{a}={-b}= {c}=k

$ Equation of plane $: \mathrm{k}(\mathrm{x}-2)- \mathrm{k}(\mathrm{y}-1)+ \mathrm{k}(\mathrm{z}+1)=0

x-y+z=0 \\

\\ $ Let the angle between $y$ -axis and plane is $ \theta$

\sin \theta=\left|\frac{0-1+0}{\sqrt{1+1+1}}\right| \\ \\ =\left|\frac{-1}{\sqrt{3}}\right| \\\\ \theta=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)

Posted by

Safeer PP

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