The angle between the planes 2x+y−2z+3=0 and 6x+3y+2z=5 is
A
cos−1(1121)
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B
cos−1(921)
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C
cos−1(711)
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D
cos−1(1329)
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Solution
The correct option is Acos−1(1121) Given planes, P1:2x+y−2z+3=0 and P2:6x+3y+2z=5 D.r′s of normal to P1:(2,1,−2) and of P2:(6,3,2)
So, cosθ=a1a2+b1b2+c1c2√a21+b21+c21⋅√a22+b22+c22 ⇒cosθ=1121⇒θ=cos−1(1121)