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Question

Find the equation of the plane that contains the line of intersection of the planes r·i^+2j^+3k^-4=0, r·2i^+j^-k^+5=0 and which is perpendicular to the plane r·5i^+3j^-6k^+8=0.

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Solution

The equation of the plane passing through the line of intersection of the given planes isr. i^+2 j^+3k^-4+λ r. 2i^+j^- k^+5=0 r. 1+2λ i^+2+λ j^+3-λk^-4+5λ=0... 1This plane is perpendicular to r. 5 i^+3 j^-6 k^+8=0. So,5 1+2λ+3 2+λ-6 3-λ=0 (Because a1a2+b1b2+c1c2=0)5 + 10λ + 6 + 3λ - 18 + 6λ = 019λ - 7 = 0λ = 719Substituting this in (1), we getr. 1 + 2 719 i ^+ 2 + 719 j ^+ 3 - 719k^ - 4 + 5 719 = 0r. 33 i ^+ 45 j^ + 50 k^ = 41

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