121384
The image of the point \((3,2,1)\) in the plane \(\mathbf{2 x}-\mathbf{y}+\mathbf{3 z}=\mathbf{7}\) is
1 \((1,2,3)\)
2 \((2,3,1)\)
3 \((3,2,1)\)
4 \((2,1,3)\)
Explanation:
C Given, Plane : \(2 \mathrm{x}-\mathrm{y}+3 \mathrm{z}=7\) Point \(\mathrm{A}(3,2,1)\) Let image of point A will be \(\mathrm{B}(\alpha, \beta, \gamma)\) So, \(\frac{\alpha-3}{2} =\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=\frac{-2(3 \times 2+2 \times(-1)+1 \times 3-7)}{4+1+9}\) \(\Rightarrow \frac{\alpha-3}{2}=\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=0\) So, \((\alpha, \beta, \gamma)=(3,2,1)\)
AP EAMCET-2009
Three Dimensional Geometry
121385
If the mid -points of the sides \(A B, B C, C A\) of a triangle are \((1,5,-1),(0,4,-2)(2,3,4)\) respectively, then length of the median drawn from \(C\) to \(A B\) is
1 \(\sqrt{29}\)
2 \(\sqrt{27}\)
3 \(\sqrt{5}\)
4 5
Explanation:
D Given, \(\mathrm{M}(1,5,-1), \mathrm{N}(0,4,-2) \text { and } \mathrm{P}(2,3,4)\) Now, using midpoint formula - \(\frac{\mathrm{x}_1+\mathrm{x}_2}{2}=1, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}=5\) And \(\frac{z_1+z_2}{2}=-1\) \(\Rightarrow \mathrm{x}_1+\mathrm{x}_2=2, \mathrm{y}_1+\mathrm{y}_2=10\) and \(\mathrm{z}_1+\mathrm{z}_2=-2\) Similarly, \(\mathrm{x}_2+\mathrm{x}_3=0, \mathrm{y}_2+\mathrm{y}_3=8 \text { and } \mathrm{z}_2+\mathrm{z}_3=-4\) And \(\mathrm{x}_1+\mathrm{x}_3=4, \mathrm{y}_1+\mathrm{y}_3=6 \text { and } \mathrm{z}_1+\mathrm{z}_3=8\) On adding all equation, \(2\left(x_1+x_2+x_3\right)=6\) \(2\left(y_1+y_2+y_3\right)=24\) \(2\left(z_1+z_2+z_3\right)=2\) and \(\Rightarrow\left(\mathrm{x}_1+\mathrm{x}_2\right)+\mathrm{x}_3=3\) \(\Rightarrow 2+\mathrm{x}_3=3\) ( \(\therefore\) from above expression) \(\Rightarrow \mathrm{x}_3=1\) Similarly, \(\mathrm{y}_3=2\) and \(\mathrm{z}_3=3\) Length of median from \(A B\) to \(C\) is - \(\mathrm{CM}=\sqrt{(1-1)^2+(2-5)^2+(-1-3)^2}\) \(\mathrm{CM}=\sqrt{25}=5\)
AP EAMCET-24.04.2018
Three Dimensional Geometry
121386
The distance of point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) with the plane \(x-y+z=5\) from the point with position vector \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) is
1 \(\sqrt{14}\)
2 \(\sqrt{42}\)
3 \(3 \sqrt{14}\)
4 \(\sqrt{3}\)
Explanation:
D Let \(\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{12}=\mathrm{k}\) \(\mathrm{x}=3 \mathrm{k}+2, \mathrm{y}=4 \mathrm{k}-1 \text { and } \mathrm{z}=12 \mathrm{k}+2\) Substituting these value in the equation of plane. \(\mathrm{x}-\mathrm{y}+\mathrm{z}=5\) \(3 \mathrm{k}+2- 4 \mathrm{k}+1+12 \mathrm{k}+2=5\) \(11 \mathrm{k}+5=5\) \(\Rightarrow \mathrm{k}=0\) \(\Rightarrow\) Point of intersection \((2,-1,2)\) Also point with position vector \((\hat{i}-2 \hat{j}+3 \hat{k})\) is \((1,-2,3)\) \(\Rightarrow\) required distance, \(d=\sqrt{(2-1)^2+(-1+2)^2+(2-3)^2}\) \(=\sqrt{1^2+1^2+1^2}=\sqrt{3}\)
AMU-2018
Three Dimensional Geometry
121388
Angle between the planes \(x+y+2 z=6\) and \(2 x\) \(-y+z=9\) is
121389
The distance of the plane \(3 x+4 y+5 z+19=0\) from the point \((1,-1,1)\) measured along a line parallel to the line with direction ratios \(2,3,1\) is
1 \(\frac{23}{5 \sqrt{2}}\)
2 \(\frac{\sqrt{71}}{5 \sqrt{2}}\)
3 \(\sqrt{14}\)
4 \(\sqrt{23}\)
Explanation:
C Given, Plane, \(P: 3 x+4 y+5 x+19=0\) \& Point \(\mathrm{A}=(1,-1,1)\) Direction ratio of parallel line is \((2,3,1)\) Equation of line passing through \(\mathrm{A}\) is - \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{1}=\lambda\) ( \(\therefore\) say) Any point on this line will be \(B=(2 \lambda+1,3 \lambda-1, \lambda+1)\) It also lies on plane \(P\). Hence, \(3(2 \lambda+1)+4(3 \lambda-1)+5(\lambda+1)+19=0\) \(\Rightarrow 23 \lambda+23=0\) \(\Rightarrow \lambda=-1\) \(\text { So, point } \mathrm{B}=(-1,-4,0)\) \(\text { Now, }\) \(\quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\Rightarrow \quad \mathrm{AB}=\sqrt{14}\) So, point \(\mathrm{B}=(-1,-4,0)\) Now, \(\Rightarrow \quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\quad \mathrm{AB}=\sqrt{14}\)
121384
The image of the point \((3,2,1)\) in the plane \(\mathbf{2 x}-\mathbf{y}+\mathbf{3 z}=\mathbf{7}\) is
1 \((1,2,3)\)
2 \((2,3,1)\)
3 \((3,2,1)\)
4 \((2,1,3)\)
Explanation:
C Given, Plane : \(2 \mathrm{x}-\mathrm{y}+3 \mathrm{z}=7\) Point \(\mathrm{A}(3,2,1)\) Let image of point A will be \(\mathrm{B}(\alpha, \beta, \gamma)\) So, \(\frac{\alpha-3}{2} =\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=\frac{-2(3 \times 2+2 \times(-1)+1 \times 3-7)}{4+1+9}\) \(\Rightarrow \frac{\alpha-3}{2}=\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=0\) So, \((\alpha, \beta, \gamma)=(3,2,1)\)
AP EAMCET-2009
Three Dimensional Geometry
121385
If the mid -points of the sides \(A B, B C, C A\) of a triangle are \((1,5,-1),(0,4,-2)(2,3,4)\) respectively, then length of the median drawn from \(C\) to \(A B\) is
1 \(\sqrt{29}\)
2 \(\sqrt{27}\)
3 \(\sqrt{5}\)
4 5
Explanation:
D Given, \(\mathrm{M}(1,5,-1), \mathrm{N}(0,4,-2) \text { and } \mathrm{P}(2,3,4)\) Now, using midpoint formula - \(\frac{\mathrm{x}_1+\mathrm{x}_2}{2}=1, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}=5\) And \(\frac{z_1+z_2}{2}=-1\) \(\Rightarrow \mathrm{x}_1+\mathrm{x}_2=2, \mathrm{y}_1+\mathrm{y}_2=10\) and \(\mathrm{z}_1+\mathrm{z}_2=-2\) Similarly, \(\mathrm{x}_2+\mathrm{x}_3=0, \mathrm{y}_2+\mathrm{y}_3=8 \text { and } \mathrm{z}_2+\mathrm{z}_3=-4\) And \(\mathrm{x}_1+\mathrm{x}_3=4, \mathrm{y}_1+\mathrm{y}_3=6 \text { and } \mathrm{z}_1+\mathrm{z}_3=8\) On adding all equation, \(2\left(x_1+x_2+x_3\right)=6\) \(2\left(y_1+y_2+y_3\right)=24\) \(2\left(z_1+z_2+z_3\right)=2\) and \(\Rightarrow\left(\mathrm{x}_1+\mathrm{x}_2\right)+\mathrm{x}_3=3\) \(\Rightarrow 2+\mathrm{x}_3=3\) ( \(\therefore\) from above expression) \(\Rightarrow \mathrm{x}_3=1\) Similarly, \(\mathrm{y}_3=2\) and \(\mathrm{z}_3=3\) Length of median from \(A B\) to \(C\) is - \(\mathrm{CM}=\sqrt{(1-1)^2+(2-5)^2+(-1-3)^2}\) \(\mathrm{CM}=\sqrt{25}=5\)
AP EAMCET-24.04.2018
Three Dimensional Geometry
121386
The distance of point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) with the plane \(x-y+z=5\) from the point with position vector \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) is
1 \(\sqrt{14}\)
2 \(\sqrt{42}\)
3 \(3 \sqrt{14}\)
4 \(\sqrt{3}\)
Explanation:
D Let \(\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{12}=\mathrm{k}\) \(\mathrm{x}=3 \mathrm{k}+2, \mathrm{y}=4 \mathrm{k}-1 \text { and } \mathrm{z}=12 \mathrm{k}+2\) Substituting these value in the equation of plane. \(\mathrm{x}-\mathrm{y}+\mathrm{z}=5\) \(3 \mathrm{k}+2- 4 \mathrm{k}+1+12 \mathrm{k}+2=5\) \(11 \mathrm{k}+5=5\) \(\Rightarrow \mathrm{k}=0\) \(\Rightarrow\) Point of intersection \((2,-1,2)\) Also point with position vector \((\hat{i}-2 \hat{j}+3 \hat{k})\) is \((1,-2,3)\) \(\Rightarrow\) required distance, \(d=\sqrt{(2-1)^2+(-1+2)^2+(2-3)^2}\) \(=\sqrt{1^2+1^2+1^2}=\sqrt{3}\)
AMU-2018
Three Dimensional Geometry
121388
Angle between the planes \(x+y+2 z=6\) and \(2 x\) \(-y+z=9\) is
121389
The distance of the plane \(3 x+4 y+5 z+19=0\) from the point \((1,-1,1)\) measured along a line parallel to the line with direction ratios \(2,3,1\) is
1 \(\frac{23}{5 \sqrt{2}}\)
2 \(\frac{\sqrt{71}}{5 \sqrt{2}}\)
3 \(\sqrt{14}\)
4 \(\sqrt{23}\)
Explanation:
C Given, Plane, \(P: 3 x+4 y+5 x+19=0\) \& Point \(\mathrm{A}=(1,-1,1)\) Direction ratio of parallel line is \((2,3,1)\) Equation of line passing through \(\mathrm{A}\) is - \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{1}=\lambda\) ( \(\therefore\) say) Any point on this line will be \(B=(2 \lambda+1,3 \lambda-1, \lambda+1)\) It also lies on plane \(P\). Hence, \(3(2 \lambda+1)+4(3 \lambda-1)+5(\lambda+1)+19=0\) \(\Rightarrow 23 \lambda+23=0\) \(\Rightarrow \lambda=-1\) \(\text { So, point } \mathrm{B}=(-1,-4,0)\) \(\text { Now, }\) \(\quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\Rightarrow \quad \mathrm{AB}=\sqrt{14}\) So, point \(\mathrm{B}=(-1,-4,0)\) Now, \(\Rightarrow \quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\quad \mathrm{AB}=\sqrt{14}\)
121384
The image of the point \((3,2,1)\) in the plane \(\mathbf{2 x}-\mathbf{y}+\mathbf{3 z}=\mathbf{7}\) is
1 \((1,2,3)\)
2 \((2,3,1)\)
3 \((3,2,1)\)
4 \((2,1,3)\)
Explanation:
C Given, Plane : \(2 \mathrm{x}-\mathrm{y}+3 \mathrm{z}=7\) Point \(\mathrm{A}(3,2,1)\) Let image of point A will be \(\mathrm{B}(\alpha, \beta, \gamma)\) So, \(\frac{\alpha-3}{2} =\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=\frac{-2(3 \times 2+2 \times(-1)+1 \times 3-7)}{4+1+9}\) \(\Rightarrow \frac{\alpha-3}{2}=\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=0\) So, \((\alpha, \beta, \gamma)=(3,2,1)\)
AP EAMCET-2009
Three Dimensional Geometry
121385
If the mid -points of the sides \(A B, B C, C A\) of a triangle are \((1,5,-1),(0,4,-2)(2,3,4)\) respectively, then length of the median drawn from \(C\) to \(A B\) is
1 \(\sqrt{29}\)
2 \(\sqrt{27}\)
3 \(\sqrt{5}\)
4 5
Explanation:
D Given, \(\mathrm{M}(1,5,-1), \mathrm{N}(0,4,-2) \text { and } \mathrm{P}(2,3,4)\) Now, using midpoint formula - \(\frac{\mathrm{x}_1+\mathrm{x}_2}{2}=1, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}=5\) And \(\frac{z_1+z_2}{2}=-1\) \(\Rightarrow \mathrm{x}_1+\mathrm{x}_2=2, \mathrm{y}_1+\mathrm{y}_2=10\) and \(\mathrm{z}_1+\mathrm{z}_2=-2\) Similarly, \(\mathrm{x}_2+\mathrm{x}_3=0, \mathrm{y}_2+\mathrm{y}_3=8 \text { and } \mathrm{z}_2+\mathrm{z}_3=-4\) And \(\mathrm{x}_1+\mathrm{x}_3=4, \mathrm{y}_1+\mathrm{y}_3=6 \text { and } \mathrm{z}_1+\mathrm{z}_3=8\) On adding all equation, \(2\left(x_1+x_2+x_3\right)=6\) \(2\left(y_1+y_2+y_3\right)=24\) \(2\left(z_1+z_2+z_3\right)=2\) and \(\Rightarrow\left(\mathrm{x}_1+\mathrm{x}_2\right)+\mathrm{x}_3=3\) \(\Rightarrow 2+\mathrm{x}_3=3\) ( \(\therefore\) from above expression) \(\Rightarrow \mathrm{x}_3=1\) Similarly, \(\mathrm{y}_3=2\) and \(\mathrm{z}_3=3\) Length of median from \(A B\) to \(C\) is - \(\mathrm{CM}=\sqrt{(1-1)^2+(2-5)^2+(-1-3)^2}\) \(\mathrm{CM}=\sqrt{25}=5\)
AP EAMCET-24.04.2018
Three Dimensional Geometry
121386
The distance of point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) with the plane \(x-y+z=5\) from the point with position vector \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) is
1 \(\sqrt{14}\)
2 \(\sqrt{42}\)
3 \(3 \sqrt{14}\)
4 \(\sqrt{3}\)
Explanation:
D Let \(\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{12}=\mathrm{k}\) \(\mathrm{x}=3 \mathrm{k}+2, \mathrm{y}=4 \mathrm{k}-1 \text { and } \mathrm{z}=12 \mathrm{k}+2\) Substituting these value in the equation of plane. \(\mathrm{x}-\mathrm{y}+\mathrm{z}=5\) \(3 \mathrm{k}+2- 4 \mathrm{k}+1+12 \mathrm{k}+2=5\) \(11 \mathrm{k}+5=5\) \(\Rightarrow \mathrm{k}=0\) \(\Rightarrow\) Point of intersection \((2,-1,2)\) Also point with position vector \((\hat{i}-2 \hat{j}+3 \hat{k})\) is \((1,-2,3)\) \(\Rightarrow\) required distance, \(d=\sqrt{(2-1)^2+(-1+2)^2+(2-3)^2}\) \(=\sqrt{1^2+1^2+1^2}=\sqrt{3}\)
AMU-2018
Three Dimensional Geometry
121388
Angle between the planes \(x+y+2 z=6\) and \(2 x\) \(-y+z=9\) is
121389
The distance of the plane \(3 x+4 y+5 z+19=0\) from the point \((1,-1,1)\) measured along a line parallel to the line with direction ratios \(2,3,1\) is
1 \(\frac{23}{5 \sqrt{2}}\)
2 \(\frac{\sqrt{71}}{5 \sqrt{2}}\)
3 \(\sqrt{14}\)
4 \(\sqrt{23}\)
Explanation:
C Given, Plane, \(P: 3 x+4 y+5 x+19=0\) \& Point \(\mathrm{A}=(1,-1,1)\) Direction ratio of parallel line is \((2,3,1)\) Equation of line passing through \(\mathrm{A}\) is - \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{1}=\lambda\) ( \(\therefore\) say) Any point on this line will be \(B=(2 \lambda+1,3 \lambda-1, \lambda+1)\) It also lies on plane \(P\). Hence, \(3(2 \lambda+1)+4(3 \lambda-1)+5(\lambda+1)+19=0\) \(\Rightarrow 23 \lambda+23=0\) \(\Rightarrow \lambda=-1\) \(\text { So, point } \mathrm{B}=(-1,-4,0)\) \(\text { Now, }\) \(\quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\Rightarrow \quad \mathrm{AB}=\sqrt{14}\) So, point \(\mathrm{B}=(-1,-4,0)\) Now, \(\Rightarrow \quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\quad \mathrm{AB}=\sqrt{14}\)
121384
The image of the point \((3,2,1)\) in the plane \(\mathbf{2 x}-\mathbf{y}+\mathbf{3 z}=\mathbf{7}\) is
1 \((1,2,3)\)
2 \((2,3,1)\)
3 \((3,2,1)\)
4 \((2,1,3)\)
Explanation:
C Given, Plane : \(2 \mathrm{x}-\mathrm{y}+3 \mathrm{z}=7\) Point \(\mathrm{A}(3,2,1)\) Let image of point A will be \(\mathrm{B}(\alpha, \beta, \gamma)\) So, \(\frac{\alpha-3}{2} =\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=\frac{-2(3 \times 2+2 \times(-1)+1 \times 3-7)}{4+1+9}\) \(\Rightarrow \frac{\alpha-3}{2}=\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=0\) So, \((\alpha, \beta, \gamma)=(3,2,1)\)
AP EAMCET-2009
Three Dimensional Geometry
121385
If the mid -points of the sides \(A B, B C, C A\) of a triangle are \((1,5,-1),(0,4,-2)(2,3,4)\) respectively, then length of the median drawn from \(C\) to \(A B\) is
1 \(\sqrt{29}\)
2 \(\sqrt{27}\)
3 \(\sqrt{5}\)
4 5
Explanation:
D Given, \(\mathrm{M}(1,5,-1), \mathrm{N}(0,4,-2) \text { and } \mathrm{P}(2,3,4)\) Now, using midpoint formula - \(\frac{\mathrm{x}_1+\mathrm{x}_2}{2}=1, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}=5\) And \(\frac{z_1+z_2}{2}=-1\) \(\Rightarrow \mathrm{x}_1+\mathrm{x}_2=2, \mathrm{y}_1+\mathrm{y}_2=10\) and \(\mathrm{z}_1+\mathrm{z}_2=-2\) Similarly, \(\mathrm{x}_2+\mathrm{x}_3=0, \mathrm{y}_2+\mathrm{y}_3=8 \text { and } \mathrm{z}_2+\mathrm{z}_3=-4\) And \(\mathrm{x}_1+\mathrm{x}_3=4, \mathrm{y}_1+\mathrm{y}_3=6 \text { and } \mathrm{z}_1+\mathrm{z}_3=8\) On adding all equation, \(2\left(x_1+x_2+x_3\right)=6\) \(2\left(y_1+y_2+y_3\right)=24\) \(2\left(z_1+z_2+z_3\right)=2\) and \(\Rightarrow\left(\mathrm{x}_1+\mathrm{x}_2\right)+\mathrm{x}_3=3\) \(\Rightarrow 2+\mathrm{x}_3=3\) ( \(\therefore\) from above expression) \(\Rightarrow \mathrm{x}_3=1\) Similarly, \(\mathrm{y}_3=2\) and \(\mathrm{z}_3=3\) Length of median from \(A B\) to \(C\) is - \(\mathrm{CM}=\sqrt{(1-1)^2+(2-5)^2+(-1-3)^2}\) \(\mathrm{CM}=\sqrt{25}=5\)
AP EAMCET-24.04.2018
Three Dimensional Geometry
121386
The distance of point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) with the plane \(x-y+z=5\) from the point with position vector \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) is
1 \(\sqrt{14}\)
2 \(\sqrt{42}\)
3 \(3 \sqrt{14}\)
4 \(\sqrt{3}\)
Explanation:
D Let \(\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{12}=\mathrm{k}\) \(\mathrm{x}=3 \mathrm{k}+2, \mathrm{y}=4 \mathrm{k}-1 \text { and } \mathrm{z}=12 \mathrm{k}+2\) Substituting these value in the equation of plane. \(\mathrm{x}-\mathrm{y}+\mathrm{z}=5\) \(3 \mathrm{k}+2- 4 \mathrm{k}+1+12 \mathrm{k}+2=5\) \(11 \mathrm{k}+5=5\) \(\Rightarrow \mathrm{k}=0\) \(\Rightarrow\) Point of intersection \((2,-1,2)\) Also point with position vector \((\hat{i}-2 \hat{j}+3 \hat{k})\) is \((1,-2,3)\) \(\Rightarrow\) required distance, \(d=\sqrt{(2-1)^2+(-1+2)^2+(2-3)^2}\) \(=\sqrt{1^2+1^2+1^2}=\sqrt{3}\)
AMU-2018
Three Dimensional Geometry
121388
Angle between the planes \(x+y+2 z=6\) and \(2 x\) \(-y+z=9\) is
121389
The distance of the plane \(3 x+4 y+5 z+19=0\) from the point \((1,-1,1)\) measured along a line parallel to the line with direction ratios \(2,3,1\) is
1 \(\frac{23}{5 \sqrt{2}}\)
2 \(\frac{\sqrt{71}}{5 \sqrt{2}}\)
3 \(\sqrt{14}\)
4 \(\sqrt{23}\)
Explanation:
C Given, Plane, \(P: 3 x+4 y+5 x+19=0\) \& Point \(\mathrm{A}=(1,-1,1)\) Direction ratio of parallel line is \((2,3,1)\) Equation of line passing through \(\mathrm{A}\) is - \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{1}=\lambda\) ( \(\therefore\) say) Any point on this line will be \(B=(2 \lambda+1,3 \lambda-1, \lambda+1)\) It also lies on plane \(P\). Hence, \(3(2 \lambda+1)+4(3 \lambda-1)+5(\lambda+1)+19=0\) \(\Rightarrow 23 \lambda+23=0\) \(\Rightarrow \lambda=-1\) \(\text { So, point } \mathrm{B}=(-1,-4,0)\) \(\text { Now, }\) \(\quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\Rightarrow \quad \mathrm{AB}=\sqrt{14}\) So, point \(\mathrm{B}=(-1,-4,0)\) Now, \(\Rightarrow \quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\quad \mathrm{AB}=\sqrt{14}\)
121384
The image of the point \((3,2,1)\) in the plane \(\mathbf{2 x}-\mathbf{y}+\mathbf{3 z}=\mathbf{7}\) is
1 \((1,2,3)\)
2 \((2,3,1)\)
3 \((3,2,1)\)
4 \((2,1,3)\)
Explanation:
C Given, Plane : \(2 \mathrm{x}-\mathrm{y}+3 \mathrm{z}=7\) Point \(\mathrm{A}(3,2,1)\) Let image of point A will be \(\mathrm{B}(\alpha, \beta, \gamma)\) So, \(\frac{\alpha-3}{2} =\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=\frac{-2(3 \times 2+2 \times(-1)+1 \times 3-7)}{4+1+9}\) \(\Rightarrow \frac{\alpha-3}{2}=\frac{\beta-2}{-1}=\frac{\gamma-1}{3}=0\) So, \((\alpha, \beta, \gamma)=(3,2,1)\)
AP EAMCET-2009
Three Dimensional Geometry
121385
If the mid -points of the sides \(A B, B C, C A\) of a triangle are \((1,5,-1),(0,4,-2)(2,3,4)\) respectively, then length of the median drawn from \(C\) to \(A B\) is
1 \(\sqrt{29}\)
2 \(\sqrt{27}\)
3 \(\sqrt{5}\)
4 5
Explanation:
D Given, \(\mathrm{M}(1,5,-1), \mathrm{N}(0,4,-2) \text { and } \mathrm{P}(2,3,4)\) Now, using midpoint formula - \(\frac{\mathrm{x}_1+\mathrm{x}_2}{2}=1, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}=5\) And \(\frac{z_1+z_2}{2}=-1\) \(\Rightarrow \mathrm{x}_1+\mathrm{x}_2=2, \mathrm{y}_1+\mathrm{y}_2=10\) and \(\mathrm{z}_1+\mathrm{z}_2=-2\) Similarly, \(\mathrm{x}_2+\mathrm{x}_3=0, \mathrm{y}_2+\mathrm{y}_3=8 \text { and } \mathrm{z}_2+\mathrm{z}_3=-4\) And \(\mathrm{x}_1+\mathrm{x}_3=4, \mathrm{y}_1+\mathrm{y}_3=6 \text { and } \mathrm{z}_1+\mathrm{z}_3=8\) On adding all equation, \(2\left(x_1+x_2+x_3\right)=6\) \(2\left(y_1+y_2+y_3\right)=24\) \(2\left(z_1+z_2+z_3\right)=2\) and \(\Rightarrow\left(\mathrm{x}_1+\mathrm{x}_2\right)+\mathrm{x}_3=3\) \(\Rightarrow 2+\mathrm{x}_3=3\) ( \(\therefore\) from above expression) \(\Rightarrow \mathrm{x}_3=1\) Similarly, \(\mathrm{y}_3=2\) and \(\mathrm{z}_3=3\) Length of median from \(A B\) to \(C\) is - \(\mathrm{CM}=\sqrt{(1-1)^2+(2-5)^2+(-1-3)^2}\) \(\mathrm{CM}=\sqrt{25}=5\)
AP EAMCET-24.04.2018
Three Dimensional Geometry
121386
The distance of point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) with the plane \(x-y+z=5\) from the point with position vector \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) is
1 \(\sqrt{14}\)
2 \(\sqrt{42}\)
3 \(3 \sqrt{14}\)
4 \(\sqrt{3}\)
Explanation:
D Let \(\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{4}=\frac{\mathrm{z}-2}{12}=\mathrm{k}\) \(\mathrm{x}=3 \mathrm{k}+2, \mathrm{y}=4 \mathrm{k}-1 \text { and } \mathrm{z}=12 \mathrm{k}+2\) Substituting these value in the equation of plane. \(\mathrm{x}-\mathrm{y}+\mathrm{z}=5\) \(3 \mathrm{k}+2- 4 \mathrm{k}+1+12 \mathrm{k}+2=5\) \(11 \mathrm{k}+5=5\) \(\Rightarrow \mathrm{k}=0\) \(\Rightarrow\) Point of intersection \((2,-1,2)\) Also point with position vector \((\hat{i}-2 \hat{j}+3 \hat{k})\) is \((1,-2,3)\) \(\Rightarrow\) required distance, \(d=\sqrt{(2-1)^2+(-1+2)^2+(2-3)^2}\) \(=\sqrt{1^2+1^2+1^2}=\sqrt{3}\)
AMU-2018
Three Dimensional Geometry
121388
Angle between the planes \(x+y+2 z=6\) and \(2 x\) \(-y+z=9\) is
121389
The distance of the plane \(3 x+4 y+5 z+19=0\) from the point \((1,-1,1)\) measured along a line parallel to the line with direction ratios \(2,3,1\) is
1 \(\frac{23}{5 \sqrt{2}}\)
2 \(\frac{\sqrt{71}}{5 \sqrt{2}}\)
3 \(\sqrt{14}\)
4 \(\sqrt{23}\)
Explanation:
C Given, Plane, \(P: 3 x+4 y+5 x+19=0\) \& Point \(\mathrm{A}=(1,-1,1)\) Direction ratio of parallel line is \((2,3,1)\) Equation of line passing through \(\mathrm{A}\) is - \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{1}=\lambda\) ( \(\therefore\) say) Any point on this line will be \(B=(2 \lambda+1,3 \lambda-1, \lambda+1)\) It also lies on plane \(P\). Hence, \(3(2 \lambda+1)+4(3 \lambda-1)+5(\lambda+1)+19=0\) \(\Rightarrow 23 \lambda+23=0\) \(\Rightarrow \lambda=-1\) \(\text { So, point } \mathrm{B}=(-1,-4,0)\) \(\text { Now, }\) \(\quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\Rightarrow \quad \mathrm{AB}=\sqrt{14}\) So, point \(\mathrm{B}=(-1,-4,0)\) Now, \(\Rightarrow \quad \mathrm{AB}=\sqrt{(-1-1)^2+(-4+1)^2+(0-1)^2}\) \(\quad \mathrm{AB}=\sqrt{14}\)
