The plane $$4x+7y+4z+81=0$$ is rotated through a right angle about its line of intersection with the plane $$5x+3y+10z=25$$. The equation of the plane in its new position is

Question

The plane $$4x+7y+4z+81=0$$ is rotated through a right  angle about its line of intersection with the plane $$5x+3y+10z=25$$. The equation of the plane in its new position is

Infinity Learn · Accepted Answer

The equation of the plane through the line of intersection of the planes $$4x$$ $$+$$ $$7y$$ $$+$$ $$4z$$ $$+$$ $$81$$ $$=$$ $$0$$ and $$5x$$ $$+$$ $$3y$$ $$+$$ $$10z$$ $$=$$ $$25$$ is $$(4x+7y+4z+81)+$$λ$$(5x+3y+10z$$−$$25)=0or$$   $$(4+5$$λ$$)x+(7+3$$λ$$)y+(4+10$$λ$$)z+81$$−$$25$$λ$$=0-----i$$ which is perpendicular to $$4x+7y+4z+81=0$$⇒$$4(4+5$$λ$$)+7(7+3$$λ$$)+4(4+10$$λ$$)=0or$$  $$81$$λ$$+81=0or$$  λ$$=$$−$$1$$ Hence, the plane is x−$$4y+6z=106$$

$The plane 4 x + 7 y + 4 z + 81 = 0 is rotated through a right angle about its line of intersection with the plane 5 x + 3 y + 10 z = 25 .$
$The equation of the plane in its new position is$

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The plane 4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane 5x+3y+10z=25. The equation of the plane in its new position is

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$The plane 4 x + 7 y + 4 z + 81 = 0 is rotated through a right angle about its line of intersection with the plane 5 x + 3 y + 10 z = 25 .$
$The equation of the plane in its new position is$