121271
The equation of the plane through intersection of planes \(x+2 y+3 z=4\) and \(2 x+y-z=-5\) and perpendicular to the plane \(5 \mathrm{x}+3 \mathrm{y}-36 \mathrm{z}=-8\)
B Given planes - \(P_1: x+2 y+3 z-4=0\) \(P_1: 2 x+y-z+5=0\) The equation of plane through intersection of \(\mathrm{P}_1\) and \(\mathrm{P}_2\) \(\text { is }\) \(P_1+\lambda P_2=0\) \((x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0\) \(x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)+(5 \lambda-4)=0\) is \(P_1+\lambda P_2=0\) \((x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0\) \(x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)+(5 \lambda-4)=0\) Since, the plane is perpendicular to \(5 \mathrm{x}+3 \mathrm{y}+6 \mathrm{z}+8 =0\) \(\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2 =0\) \((1+2 \lambda)(5)+(2+\lambda)(3)+(3-\lambda)(6)=0\) \(5+10 \lambda+6+3 \lambda+18-6 \lambda=0\) \(7 \lambda+29=0\) \(\therefore \lambda=\frac{-29}{7}\) Now, put \(\left(\lambda=\frac{-29}{7}\right)\) equation of the plane is - \((x+2 y+3 z-4)+\left(\frac{-29}{7}\right)(2 x+y-z+5)=0\) \(7 x+14 y+21 z-28-58 x-29 y+29 z-145=0\) \(-51 x-15 y+50 z-173=0\) \(51 x+15 y-50 z=-173\)
CG PET- 2016
Three Dimensional Geometry
121272
The lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}\) are
1 skew
2 parallel
3 intersecting
4 coincident
Explanation:
B Given, Line \(\left(l_1\right)=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and Line \(\left(l_2\right)=\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}\) Direction ratio of line \(\left(l_1\right)=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) is \((1,2,3)\) And, direction ratio of line \(\left(l_2\right)\) \(\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6} \text { is }\) \((-2-4,-6)=(1,2,3)\)Hence, direction ratio are equal of both given lines. So, lines are parallel.
CG PET- 2018
Three Dimensional Geometry
121273
The equation of the plane passing through the line of intersection of the planes \(x+y+z=6\) and \(2 x+3 y+4 y+4 z+5=0\) and passing through \((1,1,1)\) is
C Equations of the given planes - \(\mathrm{x}+\mathrm{y}+\mathrm{z}-6=0\) And, \(\quad 2 x+3 y+4 z+5=0\) The equation of the plane passing through the line of intersection of the plane equation. (i) and (ii) - \(x+y+z-6+\lambda(2 x+3 y+4 z+5)=0\) \(\therefore\) Plane passing through the point \((1,1,1)\) Hence, \(1+1+1-6+\lambda[2(1)+3(1)+4(1)+5]=0\) \(-3+14 \lambda=0 \quad \text { [fromEq.(iii)] }\) \(\lambda=\frac{3}{14}\) Put the value of \(\lambda\) in Eq. (iii), we get- \(x+y+z-6+\frac{3}{14}(2 x+3 y+4 z+5)=0\) \(14 x+14 y+14 z-84+6 x+9 y+12 z+15=0\) \(20 x+23 y+26 z=69\)
CG PET- 2018
Three Dimensional Geometry
121274
The equation of a sphere having centre \((1,2,3)\) and radius 3 units is \(\qquad\)
1 \(x^2+y^2+z^2-2 x-4 y-6 z=0\)
2 \(x^2+y^2+z^2-2 x-4 y-6 z+5=0\)
3 \(x^2+y^2+z^2-2 x-4 y-6 z-5=0\)
4 None of these
Explanation:
B Given, Centre \((1,2,3)\) and radius \(=3\) We know that, \(\mathrm{C}:(\mathrm{h}, \mathrm{k}, l)\), radius \(=\mathrm{r}\) Then the equation of sphere \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2+(\mathrm{z}-l)^2=\mathrm{r}^2\) So, \((1,2,3)=(\mathrm{h}, \mathrm{k}, l), \mathrm{r}=3\) \(\Rightarrow(\mathrm{x}-1)^2+(\mathrm{y}-2)^2+(\mathrm{z}-3)^2=9\) \(\Rightarrow \mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+4-4 \mathrm{y}+\mathrm{z}^2+9-6 \mathrm{z}=9\) Hence, \(x^2+y^2+z^2-2 x-4 y-6 z+5=0\)
121271
The equation of the plane through intersection of planes \(x+2 y+3 z=4\) and \(2 x+y-z=-5\) and perpendicular to the plane \(5 \mathrm{x}+3 \mathrm{y}-36 \mathrm{z}=-8\)
B Given planes - \(P_1: x+2 y+3 z-4=0\) \(P_1: 2 x+y-z+5=0\) The equation of plane through intersection of \(\mathrm{P}_1\) and \(\mathrm{P}_2\) \(\text { is }\) \(P_1+\lambda P_2=0\) \((x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0\) \(x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)+(5 \lambda-4)=0\) is \(P_1+\lambda P_2=0\) \((x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0\) \(x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)+(5 \lambda-4)=0\) Since, the plane is perpendicular to \(5 \mathrm{x}+3 \mathrm{y}+6 \mathrm{z}+8 =0\) \(\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2 =0\) \((1+2 \lambda)(5)+(2+\lambda)(3)+(3-\lambda)(6)=0\) \(5+10 \lambda+6+3 \lambda+18-6 \lambda=0\) \(7 \lambda+29=0\) \(\therefore \lambda=\frac{-29}{7}\) Now, put \(\left(\lambda=\frac{-29}{7}\right)\) equation of the plane is - \((x+2 y+3 z-4)+\left(\frac{-29}{7}\right)(2 x+y-z+5)=0\) \(7 x+14 y+21 z-28-58 x-29 y+29 z-145=0\) \(-51 x-15 y+50 z-173=0\) \(51 x+15 y-50 z=-173\)
CG PET- 2016
Three Dimensional Geometry
121272
The lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}\) are
1 skew
2 parallel
3 intersecting
4 coincident
Explanation:
B Given, Line \(\left(l_1\right)=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and Line \(\left(l_2\right)=\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}\) Direction ratio of line \(\left(l_1\right)=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) is \((1,2,3)\) And, direction ratio of line \(\left(l_2\right)\) \(\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6} \text { is }\) \((-2-4,-6)=(1,2,3)\)Hence, direction ratio are equal of both given lines. So, lines are parallel.
CG PET- 2018
Three Dimensional Geometry
121273
The equation of the plane passing through the line of intersection of the planes \(x+y+z=6\) and \(2 x+3 y+4 y+4 z+5=0\) and passing through \((1,1,1)\) is
C Equations of the given planes - \(\mathrm{x}+\mathrm{y}+\mathrm{z}-6=0\) And, \(\quad 2 x+3 y+4 z+5=0\) The equation of the plane passing through the line of intersection of the plane equation. (i) and (ii) - \(x+y+z-6+\lambda(2 x+3 y+4 z+5)=0\) \(\therefore\) Plane passing through the point \((1,1,1)\) Hence, \(1+1+1-6+\lambda[2(1)+3(1)+4(1)+5]=0\) \(-3+14 \lambda=0 \quad \text { [fromEq.(iii)] }\) \(\lambda=\frac{3}{14}\) Put the value of \(\lambda\) in Eq. (iii), we get- \(x+y+z-6+\frac{3}{14}(2 x+3 y+4 z+5)=0\) \(14 x+14 y+14 z-84+6 x+9 y+12 z+15=0\) \(20 x+23 y+26 z=69\)
CG PET- 2018
Three Dimensional Geometry
121274
The equation of a sphere having centre \((1,2,3)\) and radius 3 units is \(\qquad\)
1 \(x^2+y^2+z^2-2 x-4 y-6 z=0\)
2 \(x^2+y^2+z^2-2 x-4 y-6 z+5=0\)
3 \(x^2+y^2+z^2-2 x-4 y-6 z-5=0\)
4 None of these
Explanation:
B Given, Centre \((1,2,3)\) and radius \(=3\) We know that, \(\mathrm{C}:(\mathrm{h}, \mathrm{k}, l)\), radius \(=\mathrm{r}\) Then the equation of sphere \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2+(\mathrm{z}-l)^2=\mathrm{r}^2\) So, \((1,2,3)=(\mathrm{h}, \mathrm{k}, l), \mathrm{r}=3\) \(\Rightarrow(\mathrm{x}-1)^2+(\mathrm{y}-2)^2+(\mathrm{z}-3)^2=9\) \(\Rightarrow \mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+4-4 \mathrm{y}+\mathrm{z}^2+9-6 \mathrm{z}=9\) Hence, \(x^2+y^2+z^2-2 x-4 y-6 z+5=0\)
121271
The equation of the plane through intersection of planes \(x+2 y+3 z=4\) and \(2 x+y-z=-5\) and perpendicular to the plane \(5 \mathrm{x}+3 \mathrm{y}-36 \mathrm{z}=-8\)
B Given planes - \(P_1: x+2 y+3 z-4=0\) \(P_1: 2 x+y-z+5=0\) The equation of plane through intersection of \(\mathrm{P}_1\) and \(\mathrm{P}_2\) \(\text { is }\) \(P_1+\lambda P_2=0\) \((x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0\) \(x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)+(5 \lambda-4)=0\) is \(P_1+\lambda P_2=0\) \((x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0\) \(x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)+(5 \lambda-4)=0\) Since, the plane is perpendicular to \(5 \mathrm{x}+3 \mathrm{y}+6 \mathrm{z}+8 =0\) \(\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2 =0\) \((1+2 \lambda)(5)+(2+\lambda)(3)+(3-\lambda)(6)=0\) \(5+10 \lambda+6+3 \lambda+18-6 \lambda=0\) \(7 \lambda+29=0\) \(\therefore \lambda=\frac{-29}{7}\) Now, put \(\left(\lambda=\frac{-29}{7}\right)\) equation of the plane is - \((x+2 y+3 z-4)+\left(\frac{-29}{7}\right)(2 x+y-z+5)=0\) \(7 x+14 y+21 z-28-58 x-29 y+29 z-145=0\) \(-51 x-15 y+50 z-173=0\) \(51 x+15 y-50 z=-173\)
CG PET- 2016
Three Dimensional Geometry
121272
The lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}\) are
1 skew
2 parallel
3 intersecting
4 coincident
Explanation:
B Given, Line \(\left(l_1\right)=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and Line \(\left(l_2\right)=\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}\) Direction ratio of line \(\left(l_1\right)=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) is \((1,2,3)\) And, direction ratio of line \(\left(l_2\right)\) \(\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6} \text { is }\) \((-2-4,-6)=(1,2,3)\)Hence, direction ratio are equal of both given lines. So, lines are parallel.
CG PET- 2018
Three Dimensional Geometry
121273
The equation of the plane passing through the line of intersection of the planes \(x+y+z=6\) and \(2 x+3 y+4 y+4 z+5=0\) and passing through \((1,1,1)\) is
C Equations of the given planes - \(\mathrm{x}+\mathrm{y}+\mathrm{z}-6=0\) And, \(\quad 2 x+3 y+4 z+5=0\) The equation of the plane passing through the line of intersection of the plane equation. (i) and (ii) - \(x+y+z-6+\lambda(2 x+3 y+4 z+5)=0\) \(\therefore\) Plane passing through the point \((1,1,1)\) Hence, \(1+1+1-6+\lambda[2(1)+3(1)+4(1)+5]=0\) \(-3+14 \lambda=0 \quad \text { [fromEq.(iii)] }\) \(\lambda=\frac{3}{14}\) Put the value of \(\lambda\) in Eq. (iii), we get- \(x+y+z-6+\frac{3}{14}(2 x+3 y+4 z+5)=0\) \(14 x+14 y+14 z-84+6 x+9 y+12 z+15=0\) \(20 x+23 y+26 z=69\)
CG PET- 2018
Three Dimensional Geometry
121274
The equation of a sphere having centre \((1,2,3)\) and radius 3 units is \(\qquad\)
1 \(x^2+y^2+z^2-2 x-4 y-6 z=0\)
2 \(x^2+y^2+z^2-2 x-4 y-6 z+5=0\)
3 \(x^2+y^2+z^2-2 x-4 y-6 z-5=0\)
4 None of these
Explanation:
B Given, Centre \((1,2,3)\) and radius \(=3\) We know that, \(\mathrm{C}:(\mathrm{h}, \mathrm{k}, l)\), radius \(=\mathrm{r}\) Then the equation of sphere \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2+(\mathrm{z}-l)^2=\mathrm{r}^2\) So, \((1,2,3)=(\mathrm{h}, \mathrm{k}, l), \mathrm{r}=3\) \(\Rightarrow(\mathrm{x}-1)^2+(\mathrm{y}-2)^2+(\mathrm{z}-3)^2=9\) \(\Rightarrow \mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+4-4 \mathrm{y}+\mathrm{z}^2+9-6 \mathrm{z}=9\) Hence, \(x^2+y^2+z^2-2 x-4 y-6 z+5=0\)
121271
The equation of the plane through intersection of planes \(x+2 y+3 z=4\) and \(2 x+y-z=-5\) and perpendicular to the plane \(5 \mathrm{x}+3 \mathrm{y}-36 \mathrm{z}=-8\)
B Given planes - \(P_1: x+2 y+3 z-4=0\) \(P_1: 2 x+y-z+5=0\) The equation of plane through intersection of \(\mathrm{P}_1\) and \(\mathrm{P}_2\) \(\text { is }\) \(P_1+\lambda P_2=0\) \((x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0\) \(x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)+(5 \lambda-4)=0\) is \(P_1+\lambda P_2=0\) \((x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0\) \(x(1+2 \lambda)+y(2+\lambda)+z(3-\lambda)+(5 \lambda-4)=0\) Since, the plane is perpendicular to \(5 \mathrm{x}+3 \mathrm{y}+6 \mathrm{z}+8 =0\) \(\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2 =0\) \((1+2 \lambda)(5)+(2+\lambda)(3)+(3-\lambda)(6)=0\) \(5+10 \lambda+6+3 \lambda+18-6 \lambda=0\) \(7 \lambda+29=0\) \(\therefore \lambda=\frac{-29}{7}\) Now, put \(\left(\lambda=\frac{-29}{7}\right)\) equation of the plane is - \((x+2 y+3 z-4)+\left(\frac{-29}{7}\right)(2 x+y-z+5)=0\) \(7 x+14 y+21 z-28-58 x-29 y+29 z-145=0\) \(-51 x-15 y+50 z-173=0\) \(51 x+15 y-50 z=-173\)
CG PET- 2016
Three Dimensional Geometry
121272
The lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}\) are
1 skew
2 parallel
3 intersecting
4 coincident
Explanation:
B Given, Line \(\left(l_1\right)=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and Line \(\left(l_2\right)=\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}\) Direction ratio of line \(\left(l_1\right)=\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) is \((1,2,3)\) And, direction ratio of line \(\left(l_2\right)\) \(\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6} \text { is }\) \((-2-4,-6)=(1,2,3)\)Hence, direction ratio are equal of both given lines. So, lines are parallel.
CG PET- 2018
Three Dimensional Geometry
121273
The equation of the plane passing through the line of intersection of the planes \(x+y+z=6\) and \(2 x+3 y+4 y+4 z+5=0\) and passing through \((1,1,1)\) is
C Equations of the given planes - \(\mathrm{x}+\mathrm{y}+\mathrm{z}-6=0\) And, \(\quad 2 x+3 y+4 z+5=0\) The equation of the plane passing through the line of intersection of the plane equation. (i) and (ii) - \(x+y+z-6+\lambda(2 x+3 y+4 z+5)=0\) \(\therefore\) Plane passing through the point \((1,1,1)\) Hence, \(1+1+1-6+\lambda[2(1)+3(1)+4(1)+5]=0\) \(-3+14 \lambda=0 \quad \text { [fromEq.(iii)] }\) \(\lambda=\frac{3}{14}\) Put the value of \(\lambda\) in Eq. (iii), we get- \(x+y+z-6+\frac{3}{14}(2 x+3 y+4 z+5)=0\) \(14 x+14 y+14 z-84+6 x+9 y+12 z+15=0\) \(20 x+23 y+26 z=69\)
CG PET- 2018
Three Dimensional Geometry
121274
The equation of a sphere having centre \((1,2,3)\) and radius 3 units is \(\qquad\)
1 \(x^2+y^2+z^2-2 x-4 y-6 z=0\)
2 \(x^2+y^2+z^2-2 x-4 y-6 z+5=0\)
3 \(x^2+y^2+z^2-2 x-4 y-6 z-5=0\)
4 None of these
Explanation:
B Given, Centre \((1,2,3)\) and radius \(=3\) We know that, \(\mathrm{C}:(\mathrm{h}, \mathrm{k}, l)\), radius \(=\mathrm{r}\) Then the equation of sphere \((\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2+(\mathrm{z}-l)^2=\mathrm{r}^2\) So, \((1,2,3)=(\mathrm{h}, \mathrm{k}, l), \mathrm{r}=3\) \(\Rightarrow(\mathrm{x}-1)^2+(\mathrm{y}-2)^2+(\mathrm{z}-3)^2=9\) \(\Rightarrow \mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2+4-4 \mathrm{y}+\mathrm{z}^2+9-6 \mathrm{z}=9\) Hence, \(x^2+y^2+z^2-2 x-4 y-6 z+5=0\)
