Planes in 3D

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$The plane P_{1} : 4 x + 7 y + 4 z + 81 = 0 is rotated through a right angle about its line of$
$intersection with the plane P_{2} : 5 x + 3 y + 10 z - 25 = 0 . If P_{3} = 0 is the equation of$
$plane P_{1} in its new position and if \sqrt{k} is the distance of P_{3} = 0 from origin, then k =$

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Solution

$\begin{array}{l} Since Angle between 4 x + 7 y + 4 z + 81 = 0 and \\ (4 + 5 λ) x + (7 + 3 λ) y + (4 + 10 λ) + 81 - 25 λ = 0 is \frac{π}{2} \Rightarrow λ = - 1 \end{array}$
$\begin{array}{l} ∴ p_{3} \equiv x - 4 y + 6 z - 106 = 0 \\ Since the distance from (0, 0) to P_{3} is \sqrt{k} \\ ∴ \sqrt{k} = \frac{106}{\sqrt{53}} = 2 \sqrt{53} \\ \Rightarrow k = 212 \end{array}$

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Planes in 3D

Remember concepts with our Masterclasses.

The plane P1:4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane P2:5x+3y+10z−25=0 . If P3=0 is the equation of plane P1 in its new position and if k is the distance of P3=0 from origin, then k=

Since Angle between 4x+7y+4z+81=0 and (4+5λ)x+(7+3λ)y+(4+10λ)+81−25λ=0 is π2⇒λ=−1∴p3≡x−4y+6z−106=0 Since the distance from (0,0) to P3 is k∴k=10653=253⇒k=212

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$The plane P_{1} : 4 x + 7 y + 4 z + 81 = 0 is rotated through a right angle about its line of$
$intersection with the plane P_{2} : 5 x + 3 y + 10 z - 25 = 0 . If P_{3} = 0 is the equation of$
$plane P_{1} in its new position and if \sqrt{k} is the distance of P_{3} = 0 from origin, then k =$