Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of a plane $P$ passing through the line of intersection of $P_1:x+2y-z-1=0$ and $P_2:2x-y+z-2=0$ is given by $P_1+\lambda P_2=0$.$$(x+2y-z-1)+\lambda (2x-y+z-2)=0$$ $$(1+2\lambda )x+(2-\lambda )y+(-1+\lambda )z-(1+2\lambda )=0$$ The normal vector to this plane is $\vec{n}=(1+2\lambda ,2-\lambda ,-1+\lambda )$. **(P) Plane passes through $(1,0,1)$:** Substituting $(1,0,1)$ into the plane equation: $(1+2\lambda )(1)+(2-\lambda )(0)+(-1+\lambda )(1)-(1+2\lambda )=0⟹1+2\lambda -1+\lambda -1-2\lambda =0⟹\lambda -1=0⟹\lambda =1$. The plane $P_P$ is $3x+y-3=0$. For List-II (1), check if $(0,3,1)$ lies on $P_P$: $3(0)+3-3=0$. Yes. So (P)-(1). **(Q) Plane is perpendicular to $x+y+z=0$:** The normal vector of $x+y+z=0$ is $(1,1,1)$. For perpendicularity, $\vec{n}\cdot (1,1,1)=0$.$(1+2\lambda )(1)+(2-\lambda )(1)+(-1+\lambda )(1)=0⟹1+2\lambda +2-\lambda -1+\lambda =0⟹2+2\lambda =0⟹\lambda =-1$.

The plane $P_Q$ is $-x+3y-2z+1=0$, or $x-3y+2z-1=0$. For List-II (3), the angle with the $xy$ -plane (normal $(0,0,1)$) is ${\cos }^{-1}\left(\frac{|(1,-3,2)\cdot (0,0,1)|}{\Vert (1,-3,2)\Vert \Vert (0,0,1)\Vert }\right)={\cos }^{-1}\left(\frac{|2|}{\sqrt{1^2+(-3{)}^2+2^2}}\right)={\cos }^{-1}\left(\frac{2}{\sqrt{14}}\right)$. So (Q)-(3). **(R) Plane is parallel to the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-2}{-1}$:** The direction vector of the line is $(2,1,-1)$. For parallelism, $\vec{n}\cdot (2,1,-1)=0$.$(1+2\lambda )(2)+(2-\lambda )(1)+(-1+\lambda )(-1)=0⟹2+4\lambda +2-\lambda +1-\lambda =0⟹5+2\lambda =0⟹\lambda =-\frac{5}{2}$. The plane $P_R$ is $(1-5)x+(2+\frac{5}{2})y+(-1-\frac{5}{2})z-(1-5)=0⟹-4x+\frac{9}{2}y-\frac{7}{2}z+4=0⟹8x-9y+7z-8=0$. For List-II (4), the angle with the $xy$ -plane is ${\cos }^{-1}\left(\frac{|(8,-9,7)\cdot (0,0,1)|}{\Vert (8,-9,7)\Vert \Vert (0,0,1)\Vert }\right)={\cos }^{-1}\left(\frac{|7|}{\sqrt{8^2+(-9{)}^2+7^2}}\right)={\cos }^{-1}\left(\frac{7}{\sqrt{194}}\right)$. So (R)-(4). **(S) Plane at distance $\frac{1}{\sqrt{6}}$ from origin and normal's $y$ -component positive:** Distance from origin: $\frac{|-(1+2\lambda )|}{\sqrt{(1+2\lambda {)}^2+(2-\lambda {)}^2+(-1+\lambda {)}^2}}=\frac{1}{\sqrt{6}}$.$\frac{|1+2\lambda |}{\sqrt{6{\lambda }^2-2\lambda +6}}=\frac{1}{\sqrt{6}}⟹6(1+2\lambda {)}^2=6{\lambda }^2-2\lambda +6⟹6(1+4\lambda +4{\lambda }^2)=6{\lambda }^2-2\lambda +6⟹18{\lambda }^2+26\lambda =0⟹2\lambda (9\lambda +13)=0$. So, $\lambda =0$ or $\lambda =-\frac{13}{9}$.

If $\lambda =0$, the plane is $x+2y-z-1=0$. Its normal vector is $(1,2,-1)$, $y$ -component is $2$ (positive). This is $P_S$.

If $\lambda =-\frac{13}{9}$, the plane is $17x-31y+22z-17=0$. Its normal vector is $(17,-31,22)$, $y$ -component is $-31$ (negative). This plane is rejected. So, the unique plane for (S) is $x+2y-z-1=0$. For List-II (2), the angle with the $xy$ -plane is ${\cos }^{-1}\left(\frac{|(1,2,-1)\cdot (0,0,1)|}{\Vert (1,2,-1)\Vert \Vert (0,0,1)\Vert }\right)={\cos }^{-1}\left(\frac{|-1|}{\sqrt{1^2+2^2+(-1{)}^2}}\right)={\cos }^{-1}\left(\frac{1}{\sqrt{6}}\right)$. So (S)-(2). Thus, the correct matching is (P)-(1), (Q)-(3), (R)-(4), (S)-(2).