Let \(P\) be the plane such that it contains the straight line $$\frac{x-1}{2}=\frac{y-3}{3}=\frac{z+2}{1}$$ and is perpendicular to the plane $$x+2y+3z=4.$$ Let \(P_1\) be the plane which passes through the point \((4,2,2)\) and is parallel to \(P\). Then which of the following statements is (are) TRUE?
The line has direction vector \((2,3,1)\). The given plane has normal vector \((1,2,3)\). Since plane P contains the line and is perpendicular to the given plane, a normal vector to P is \((2,3,1)\times(1,2,3)=(7,-5,1)\). Using point \((1,3,-2)\) on the line, the equation of P is \(7x-5y+z=-10\). Hence (A) is true. The distance between P and the parallel plane through \((4,2,2)\) is \(\frac{|7(4)-5(2)+2+10|}{\sqrt{75}}=\frac{30}{5\sqrt3}=2\sqrt3\), so (B) is false. The distance of P from the origin is \(\frac{10}{\sqrt{75}}=\frac{2}{\sqrt3}\), so (C) is false. The angle between planes equals the angle between their normals. Normals are \((7,-5,1)\) and \((2,2,1)\). Therefore \(\cos\theta=\frac{|14-10+1|}{\sqrt{75}\cdot3}=\frac{5}{15\sqrt3}=\frac1{3\sqrt3}\), so (D) is true.
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