121281
The equation of the plane through the intersection of the planes \(x+2 y+3 z-4=0\) and \(4 x+3 y+2 z+1=0\) and passing through the origin is ........
1 \(17+14 y+11 z=0\)
2 \(7 x+4 y+z=0\)
3 \(x+14 y+11 z=0\)
4 \(17 x+y+z=0\)
Explanation:
A Given, plane : \(\mathrm{P}_1: \mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4=0\) \(P_2: 4 x+3 y+2 x+1=0\) Equation of required plane - \(\mathrm{P}=\mathrm{P}_1+\lambda \mathrm{P}_2\) \(\mathrm{P}=(\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4)+\lambda(4 \mathrm{x}+3 \mathrm{y}+2 \mathrm{x}+1)=0\) Also, it passes through origin \((0,0,0)\) So, \(-4+\lambda=0\) Hence, \((x+2 y+3 z-4)+4(4 x+3 y+2 z+1)=0\) \(17 x+14 y+11 z=0\)
AP EAMCET-18.09.2020
Three Dimensional Geometry
121282
If \((2,3,-3)\) is one end of a diameter of the sphere \(x^2+y^2+z^2-6 x-12 y-2 z+20=0\), then the other end of the diameter is
1 \((4,9,-1)\)
2 \((4,9,5)\)
3 \((-8,-15,1)\)
4 \((8,15,5)\)
Explanation:
B Equation of sphere, \(x^2+y^2+z^2-6 x-12 y-2 z+20=0\) One end of diameter is \((2,3,-3)\) Coordinate of centre of sphere \(=(3,6,1)\) Let coordinate of other will be \((\alpha, \beta, \gamma)\). Then, \(\frac{\alpha+2}{2}=3 \Rightarrow \alpha=4\) \(\frac{\beta+3}{2}=6 \Rightarrow \beta=9\) \(\frac{\gamma+(-3)}{2}=1 \Rightarrow \gamma=5\)Hence, \((\alpha, \beta, \gamma)=(4,9,5)\)
AP EAMCET-2010
Three Dimensional Geometry
121283
The radius of the sphere \(x^2+y^2+z^2=12 x+4 y+3 z\) is
1 \(\frac{13}{2}\)
2 13
3 26
4 52
Explanation:
A Given sphere: \(x^2+y^2+z^2=12 x+4 y+3 z\) \(\left[x^2-12 x+(6)^2\right]-(6)^2+\left[y^2-4 y+(2)^2\right]-(2)^2\) \(+\left[z^2-3 z+\left(\frac{3}{2}\right)^2\right]-\left(\frac{3}{2}\right)^2=0\) \((x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2=36+4+\frac{9}{4}\) \((x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2=\frac{169}{4}=\left(\frac{13}{2}\right)^2\) \(\text { On comparing this with - }\) \((x-a)^2+(y-b)^2+(z-c)^2=r^2\) \(\text { Radius of sphere }(r)=\frac{13}{2}\) On comparing this with - Radius of sphere \((r)=\frac{13}{2}\)
AP EAMCET-2009
Three Dimensional Geometry
121284
The equation of the plane bisecting the line segment joining the points \((2,0,6)\) and \((-6,2\), 4) and perpendicular to it, is
1 \(2 x-y+4 z-15=0\)
2 \(4 x-y+3 z-6=0\)
3 \(4 x-y+z+4=0\)
4 \(x-2 y+3 z-11=0\)
Explanation:
C Given, points A \((2,0,6)\) and B \((-6,2,4)\) Equation of required plane bisecting the line segment \(\mathrm{AB}\) can be calculated as - \(\mathrm{AB}=\left(\frac{2-6}{2}, \frac{0+2}{2}, \frac{6+4}{2}\right)\) \(\mathrm{AB}=(-2,1,5) \quad \text { (bisector point) }\) \(\overrightarrow{\mathrm{n}}=(-6-2) \hat{\mathrm{i}}+(2-0) \hat{\mathrm{j}}+(4-6) \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{n}}=-8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) Follows relation, \(-8 \mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=-8(-2)+2(1)-2(5)=8\) Hence, equation of plane - \(4 \mathrm{x}-\mathrm{y}+\mathrm{z}+4=0\)
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Three Dimensional Geometry
121281
The equation of the plane through the intersection of the planes \(x+2 y+3 z-4=0\) and \(4 x+3 y+2 z+1=0\) and passing through the origin is ........
1 \(17+14 y+11 z=0\)
2 \(7 x+4 y+z=0\)
3 \(x+14 y+11 z=0\)
4 \(17 x+y+z=0\)
Explanation:
A Given, plane : \(\mathrm{P}_1: \mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4=0\) \(P_2: 4 x+3 y+2 x+1=0\) Equation of required plane - \(\mathrm{P}=\mathrm{P}_1+\lambda \mathrm{P}_2\) \(\mathrm{P}=(\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4)+\lambda(4 \mathrm{x}+3 \mathrm{y}+2 \mathrm{x}+1)=0\) Also, it passes through origin \((0,0,0)\) So, \(-4+\lambda=0\) Hence, \((x+2 y+3 z-4)+4(4 x+3 y+2 z+1)=0\) \(17 x+14 y+11 z=0\)
AP EAMCET-18.09.2020
Three Dimensional Geometry
121282
If \((2,3,-3)\) is one end of a diameter of the sphere \(x^2+y^2+z^2-6 x-12 y-2 z+20=0\), then the other end of the diameter is
1 \((4,9,-1)\)
2 \((4,9,5)\)
3 \((-8,-15,1)\)
4 \((8,15,5)\)
Explanation:
B Equation of sphere, \(x^2+y^2+z^2-6 x-12 y-2 z+20=0\) One end of diameter is \((2,3,-3)\) Coordinate of centre of sphere \(=(3,6,1)\) Let coordinate of other will be \((\alpha, \beta, \gamma)\). Then, \(\frac{\alpha+2}{2}=3 \Rightarrow \alpha=4\) \(\frac{\beta+3}{2}=6 \Rightarrow \beta=9\) \(\frac{\gamma+(-3)}{2}=1 \Rightarrow \gamma=5\)Hence, \((\alpha, \beta, \gamma)=(4,9,5)\)
AP EAMCET-2010
Three Dimensional Geometry
121283
The radius of the sphere \(x^2+y^2+z^2=12 x+4 y+3 z\) is
1 \(\frac{13}{2}\)
2 13
3 26
4 52
Explanation:
A Given sphere: \(x^2+y^2+z^2=12 x+4 y+3 z\) \(\left[x^2-12 x+(6)^2\right]-(6)^2+\left[y^2-4 y+(2)^2\right]-(2)^2\) \(+\left[z^2-3 z+\left(\frac{3}{2}\right)^2\right]-\left(\frac{3}{2}\right)^2=0\) \((x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2=36+4+\frac{9}{4}\) \((x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2=\frac{169}{4}=\left(\frac{13}{2}\right)^2\) \(\text { On comparing this with - }\) \((x-a)^2+(y-b)^2+(z-c)^2=r^2\) \(\text { Radius of sphere }(r)=\frac{13}{2}\) On comparing this with - Radius of sphere \((r)=\frac{13}{2}\)
AP EAMCET-2009
Three Dimensional Geometry
121284
The equation of the plane bisecting the line segment joining the points \((2,0,6)\) and \((-6,2\), 4) and perpendicular to it, is
1 \(2 x-y+4 z-15=0\)
2 \(4 x-y+3 z-6=0\)
3 \(4 x-y+z+4=0\)
4 \(x-2 y+3 z-11=0\)
Explanation:
C Given, points A \((2,0,6)\) and B \((-6,2,4)\) Equation of required plane bisecting the line segment \(\mathrm{AB}\) can be calculated as - \(\mathrm{AB}=\left(\frac{2-6}{2}, \frac{0+2}{2}, \frac{6+4}{2}\right)\) \(\mathrm{AB}=(-2,1,5) \quad \text { (bisector point) }\) \(\overrightarrow{\mathrm{n}}=(-6-2) \hat{\mathrm{i}}+(2-0) \hat{\mathrm{j}}+(4-6) \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{n}}=-8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) Follows relation, \(-8 \mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=-8(-2)+2(1)-2(5)=8\) Hence, equation of plane - \(4 \mathrm{x}-\mathrm{y}+\mathrm{z}+4=0\)
121281
The equation of the plane through the intersection of the planes \(x+2 y+3 z-4=0\) and \(4 x+3 y+2 z+1=0\) and passing through the origin is ........
1 \(17+14 y+11 z=0\)
2 \(7 x+4 y+z=0\)
3 \(x+14 y+11 z=0\)
4 \(17 x+y+z=0\)
Explanation:
A Given, plane : \(\mathrm{P}_1: \mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4=0\) \(P_2: 4 x+3 y+2 x+1=0\) Equation of required plane - \(\mathrm{P}=\mathrm{P}_1+\lambda \mathrm{P}_2\) \(\mathrm{P}=(\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4)+\lambda(4 \mathrm{x}+3 \mathrm{y}+2 \mathrm{x}+1)=0\) Also, it passes through origin \((0,0,0)\) So, \(-4+\lambda=0\) Hence, \((x+2 y+3 z-4)+4(4 x+3 y+2 z+1)=0\) \(17 x+14 y+11 z=0\)
AP EAMCET-18.09.2020
Three Dimensional Geometry
121282
If \((2,3,-3)\) is one end of a diameter of the sphere \(x^2+y^2+z^2-6 x-12 y-2 z+20=0\), then the other end of the diameter is
1 \((4,9,-1)\)
2 \((4,9,5)\)
3 \((-8,-15,1)\)
4 \((8,15,5)\)
Explanation:
B Equation of sphere, \(x^2+y^2+z^2-6 x-12 y-2 z+20=0\) One end of diameter is \((2,3,-3)\) Coordinate of centre of sphere \(=(3,6,1)\) Let coordinate of other will be \((\alpha, \beta, \gamma)\). Then, \(\frac{\alpha+2}{2}=3 \Rightarrow \alpha=4\) \(\frac{\beta+3}{2}=6 \Rightarrow \beta=9\) \(\frac{\gamma+(-3)}{2}=1 \Rightarrow \gamma=5\)Hence, \((\alpha, \beta, \gamma)=(4,9,5)\)
AP EAMCET-2010
Three Dimensional Geometry
121283
The radius of the sphere \(x^2+y^2+z^2=12 x+4 y+3 z\) is
1 \(\frac{13}{2}\)
2 13
3 26
4 52
Explanation:
A Given sphere: \(x^2+y^2+z^2=12 x+4 y+3 z\) \(\left[x^2-12 x+(6)^2\right]-(6)^2+\left[y^2-4 y+(2)^2\right]-(2)^2\) \(+\left[z^2-3 z+\left(\frac{3}{2}\right)^2\right]-\left(\frac{3}{2}\right)^2=0\) \((x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2=36+4+\frac{9}{4}\) \((x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2=\frac{169}{4}=\left(\frac{13}{2}\right)^2\) \(\text { On comparing this with - }\) \((x-a)^2+(y-b)^2+(z-c)^2=r^2\) \(\text { Radius of sphere }(r)=\frac{13}{2}\) On comparing this with - Radius of sphere \((r)=\frac{13}{2}\)
AP EAMCET-2009
Three Dimensional Geometry
121284
The equation of the plane bisecting the line segment joining the points \((2,0,6)\) and \((-6,2\), 4) and perpendicular to it, is
1 \(2 x-y+4 z-15=0\)
2 \(4 x-y+3 z-6=0\)
3 \(4 x-y+z+4=0\)
4 \(x-2 y+3 z-11=0\)
Explanation:
C Given, points A \((2,0,6)\) and B \((-6,2,4)\) Equation of required plane bisecting the line segment \(\mathrm{AB}\) can be calculated as - \(\mathrm{AB}=\left(\frac{2-6}{2}, \frac{0+2}{2}, \frac{6+4}{2}\right)\) \(\mathrm{AB}=(-2,1,5) \quad \text { (bisector point) }\) \(\overrightarrow{\mathrm{n}}=(-6-2) \hat{\mathrm{i}}+(2-0) \hat{\mathrm{j}}+(4-6) \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{n}}=-8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) Follows relation, \(-8 \mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=-8(-2)+2(1)-2(5)=8\) Hence, equation of plane - \(4 \mathrm{x}-\mathrm{y}+\mathrm{z}+4=0\)
121281
The equation of the plane through the intersection of the planes \(x+2 y+3 z-4=0\) and \(4 x+3 y+2 z+1=0\) and passing through the origin is ........
1 \(17+14 y+11 z=0\)
2 \(7 x+4 y+z=0\)
3 \(x+14 y+11 z=0\)
4 \(17 x+y+z=0\)
Explanation:
A Given, plane : \(\mathrm{P}_1: \mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4=0\) \(P_2: 4 x+3 y+2 x+1=0\) Equation of required plane - \(\mathrm{P}=\mathrm{P}_1+\lambda \mathrm{P}_2\) \(\mathrm{P}=(\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}-4)+\lambda(4 \mathrm{x}+3 \mathrm{y}+2 \mathrm{x}+1)=0\) Also, it passes through origin \((0,0,0)\) So, \(-4+\lambda=0\) Hence, \((x+2 y+3 z-4)+4(4 x+3 y+2 z+1)=0\) \(17 x+14 y+11 z=0\)
AP EAMCET-18.09.2020
Three Dimensional Geometry
121282
If \((2,3,-3)\) is one end of a diameter of the sphere \(x^2+y^2+z^2-6 x-12 y-2 z+20=0\), then the other end of the diameter is
1 \((4,9,-1)\)
2 \((4,9,5)\)
3 \((-8,-15,1)\)
4 \((8,15,5)\)
Explanation:
B Equation of sphere, \(x^2+y^2+z^2-6 x-12 y-2 z+20=0\) One end of diameter is \((2,3,-3)\) Coordinate of centre of sphere \(=(3,6,1)\) Let coordinate of other will be \((\alpha, \beta, \gamma)\). Then, \(\frac{\alpha+2}{2}=3 \Rightarrow \alpha=4\) \(\frac{\beta+3}{2}=6 \Rightarrow \beta=9\) \(\frac{\gamma+(-3)}{2}=1 \Rightarrow \gamma=5\)Hence, \((\alpha, \beta, \gamma)=(4,9,5)\)
AP EAMCET-2010
Three Dimensional Geometry
121283
The radius of the sphere \(x^2+y^2+z^2=12 x+4 y+3 z\) is
1 \(\frac{13}{2}\)
2 13
3 26
4 52
Explanation:
A Given sphere: \(x^2+y^2+z^2=12 x+4 y+3 z\) \(\left[x^2-12 x+(6)^2\right]-(6)^2+\left[y^2-4 y+(2)^2\right]-(2)^2\) \(+\left[z^2-3 z+\left(\frac{3}{2}\right)^2\right]-\left(\frac{3}{2}\right)^2=0\) \((x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2=36+4+\frac{9}{4}\) \((x-6)^2+(y-2)^2+\left(z-\frac{3}{2}\right)^2=\frac{169}{4}=\left(\frac{13}{2}\right)^2\) \(\text { On comparing this with - }\) \((x-a)^2+(y-b)^2+(z-c)^2=r^2\) \(\text { Radius of sphere }(r)=\frac{13}{2}\) On comparing this with - Radius of sphere \((r)=\frac{13}{2}\)
AP EAMCET-2009
Three Dimensional Geometry
121284
The equation of the plane bisecting the line segment joining the points \((2,0,6)\) and \((-6,2\), 4) and perpendicular to it, is
1 \(2 x-y+4 z-15=0\)
2 \(4 x-y+3 z-6=0\)
3 \(4 x-y+z+4=0\)
4 \(x-2 y+3 z-11=0\)
Explanation:
C Given, points A \((2,0,6)\) and B \((-6,2,4)\) Equation of required plane bisecting the line segment \(\mathrm{AB}\) can be calculated as - \(\mathrm{AB}=\left(\frac{2-6}{2}, \frac{0+2}{2}, \frac{6+4}{2}\right)\) \(\mathrm{AB}=(-2,1,5) \quad \text { (bisector point) }\) \(\overrightarrow{\mathrm{n}}=(-6-2) \hat{\mathrm{i}}+(2-0) \hat{\mathrm{j}}+(4-6) \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{n}}=-8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) Follows relation, \(-8 \mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=-8(-2)+2(1)-2(5)=8\) Hence, equation of plane - \(4 \mathrm{x}-\mathrm{y}+\mathrm{z}+4=0\)