88438
The locus of the point, which moves such that its distance from \((1,-2,2)\) is unity, will be
1 \(x^{2}+y^{2}+z^{2}-2 x+4 y-4 z+8=0\)
2 \(x^{2}+y^{2}+z^{2}-2 x-4 y-4 z+8=0\)
3 \(x^{2}+y^{2}+z^{2}+2 x+4 y-4 z+8=0\)
4 \(x^{2}+y^{2}+z^{2}-2 x+4 y+4 z+8=0\)
Explanation:
(A) Let, The point in \(\mathrm{P}=(1,-2,2)\) So, \(\quad \mathrm{Q}=(\mathrm{x}, \mathrm{y}, \mathrm{z})\) Given \((\mathrm{PQ})=1\) \(\text { So, } \quad \sqrt{(x-1)^2+(y+2)^2+(z-2)^2}=1\) Square both side \((x-1)^{2}+(y+2)^{2}+(z-2)^{2}=1\) \(x^{2}-2 x+1+y^{2}+4 y+4+z^{2}-4 z+4=1\) Therefore locus of point \(\mathrm{P}\) is \(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-2 \mathrm{x}+4 \mathrm{y}-4 \mathrm{z}+8=0\)
CG PET-2010
Co-Ordinate system
88439
A variable plane is at a constant distance \(P\) from the origin \(O\) and meets the axes at \(A, B\) and \(C\). The locus of the centroid of the tetrahedron \(\mathrm{OABC}\) is
(B) : Let the plane meets then axes at \(\mathrm{A}(\mathrm{a}, 0,0)\), \(B(0, b, 0)\) and \(C(0,0, c)\) then, equation of plane is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\) It is given that plane is at a constant distance \(\mathrm{p}\) from the origin \((0,0,0)\) \(\therefore \quad \mathrm{p}=\left|\frac{0+0-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}}}\right|\) \(\Rightarrow \quad \frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}\) Let \((\alpha, \beta, \gamma)\) be the coordinates of centroid of the formed tetrahedron. Then, \(\alpha=\frac{a+0+0+0}{4} \Rightarrow a=4 \alpha\) \(\beta=\frac{0+b+0+0}{4} \Rightarrow b=4 \beta\) \(\gamma=\frac{0+0+c+0}{4} \Rightarrow c=4 \gamma\) On putting values of \(a b, c\) in Eq. (i) we get \(\frac{1}{\mathrm{p}^{2}}=\frac{1}{(4 \alpha)^{2}}+\frac{1}{(4 \beta)^{2}}+\frac{1}{(4 \gamma)^{2}}\) \(\Rightarrow \frac{1}{\mathrm{p}^{2}}=\frac{1}{16 \alpha^{2}}+\frac{1}{16 \beta^{2}}+\frac{1}{16 \gamma^{2}}\) \(\Rightarrow \frac{16}{p^{2}}=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}\) \(\therefore\) Locus of centroid of tetrahedron is \(\frac{16}{\mathrm{p}^{2}}=\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}+\frac{1}{\mathrm{z}^{2}}\)
CG PET-2015
Co-Ordinate system
88440
A line segment \(A B\) of length \(\lambda\) moves such that the points \(A\) and \(B\) remain on the periphery of a circle of radius \(\lambda\). Then the locus of the point, that divides the line segment \(A B\) in the ratio 2: 3 is a circle of radius
88441
A rod of length \(2 l\) slides with its ends on two perpendicular lines, then the locus of its midpoint is
1 \(\mathrm{x}^{2}+\mathrm{y}^{2}=l^{2}\)
2 \(\mathrm{x}^{2}-\mathrm{y}^{2}=l^{2}\)
3 \(2 \mathrm{x}^{2}+2 \mathrm{y}^{2}=l^{2}\)
4 \(2 x^{2}-2 y^{2}=l^{2}\)
Explanation:
(A): Let \(\mathrm{D}\) be the mid point on rod. Let \(\mathrm{x}=2 \alpha\) and \(\mathrm{y}=2 \beta\) \(\because x =2 \alpha \Rightarrow \alpha=\frac{x}{2}\) \(y =2 \beta \Rightarrow \beta=\frac{y}{2}\) In \(\triangle \mathrm{AOB}\), \(\mathrm{x}^{2}+\mathrm{y}^{2}=(2 l)^{2}\) \((2 \alpha)^{2}+(2 \beta)^{2}=4 l^{2}\) \(4 \alpha^{2}+4 \beta^{2}=4 l^{2}\) \(\alpha^{2}+\beta^{2}=l^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}=l^{2}\)
88438
The locus of the point, which moves such that its distance from \((1,-2,2)\) is unity, will be
1 \(x^{2}+y^{2}+z^{2}-2 x+4 y-4 z+8=0\)
2 \(x^{2}+y^{2}+z^{2}-2 x-4 y-4 z+8=0\)
3 \(x^{2}+y^{2}+z^{2}+2 x+4 y-4 z+8=0\)
4 \(x^{2}+y^{2}+z^{2}-2 x+4 y+4 z+8=0\)
Explanation:
(A) Let, The point in \(\mathrm{P}=(1,-2,2)\) So, \(\quad \mathrm{Q}=(\mathrm{x}, \mathrm{y}, \mathrm{z})\) Given \((\mathrm{PQ})=1\) \(\text { So, } \quad \sqrt{(x-1)^2+(y+2)^2+(z-2)^2}=1\) Square both side \((x-1)^{2}+(y+2)^{2}+(z-2)^{2}=1\) \(x^{2}-2 x+1+y^{2}+4 y+4+z^{2}-4 z+4=1\) Therefore locus of point \(\mathrm{P}\) is \(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-2 \mathrm{x}+4 \mathrm{y}-4 \mathrm{z}+8=0\)
CG PET-2010
Co-Ordinate system
88439
A variable plane is at a constant distance \(P\) from the origin \(O\) and meets the axes at \(A, B\) and \(C\). The locus of the centroid of the tetrahedron \(\mathrm{OABC}\) is
(B) : Let the plane meets then axes at \(\mathrm{A}(\mathrm{a}, 0,0)\), \(B(0, b, 0)\) and \(C(0,0, c)\) then, equation of plane is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\) It is given that plane is at a constant distance \(\mathrm{p}\) from the origin \((0,0,0)\) \(\therefore \quad \mathrm{p}=\left|\frac{0+0-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}}}\right|\) \(\Rightarrow \quad \frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}\) Let \((\alpha, \beta, \gamma)\) be the coordinates of centroid of the formed tetrahedron. Then, \(\alpha=\frac{a+0+0+0}{4} \Rightarrow a=4 \alpha\) \(\beta=\frac{0+b+0+0}{4} \Rightarrow b=4 \beta\) \(\gamma=\frac{0+0+c+0}{4} \Rightarrow c=4 \gamma\) On putting values of \(a b, c\) in Eq. (i) we get \(\frac{1}{\mathrm{p}^{2}}=\frac{1}{(4 \alpha)^{2}}+\frac{1}{(4 \beta)^{2}}+\frac{1}{(4 \gamma)^{2}}\) \(\Rightarrow \frac{1}{\mathrm{p}^{2}}=\frac{1}{16 \alpha^{2}}+\frac{1}{16 \beta^{2}}+\frac{1}{16 \gamma^{2}}\) \(\Rightarrow \frac{16}{p^{2}}=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}\) \(\therefore\) Locus of centroid of tetrahedron is \(\frac{16}{\mathrm{p}^{2}}=\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}+\frac{1}{\mathrm{z}^{2}}\)
CG PET-2015
Co-Ordinate system
88440
A line segment \(A B\) of length \(\lambda\) moves such that the points \(A\) and \(B\) remain on the periphery of a circle of radius \(\lambda\). Then the locus of the point, that divides the line segment \(A B\) in the ratio 2: 3 is a circle of radius
88441
A rod of length \(2 l\) slides with its ends on two perpendicular lines, then the locus of its midpoint is
1 \(\mathrm{x}^{2}+\mathrm{y}^{2}=l^{2}\)
2 \(\mathrm{x}^{2}-\mathrm{y}^{2}=l^{2}\)
3 \(2 \mathrm{x}^{2}+2 \mathrm{y}^{2}=l^{2}\)
4 \(2 x^{2}-2 y^{2}=l^{2}\)
Explanation:
(A): Let \(\mathrm{D}\) be the mid point on rod. Let \(\mathrm{x}=2 \alpha\) and \(\mathrm{y}=2 \beta\) \(\because x =2 \alpha \Rightarrow \alpha=\frac{x}{2}\) \(y =2 \beta \Rightarrow \beta=\frac{y}{2}\) In \(\triangle \mathrm{AOB}\), \(\mathrm{x}^{2}+\mathrm{y}^{2}=(2 l)^{2}\) \((2 \alpha)^{2}+(2 \beta)^{2}=4 l^{2}\) \(4 \alpha^{2}+4 \beta^{2}=4 l^{2}\) \(\alpha^{2}+\beta^{2}=l^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}=l^{2}\)
88438
The locus of the point, which moves such that its distance from \((1,-2,2)\) is unity, will be
1 \(x^{2}+y^{2}+z^{2}-2 x+4 y-4 z+8=0\)
2 \(x^{2}+y^{2}+z^{2}-2 x-4 y-4 z+8=0\)
3 \(x^{2}+y^{2}+z^{2}+2 x+4 y-4 z+8=0\)
4 \(x^{2}+y^{2}+z^{2}-2 x+4 y+4 z+8=0\)
Explanation:
(A) Let, The point in \(\mathrm{P}=(1,-2,2)\) So, \(\quad \mathrm{Q}=(\mathrm{x}, \mathrm{y}, \mathrm{z})\) Given \((\mathrm{PQ})=1\) \(\text { So, } \quad \sqrt{(x-1)^2+(y+2)^2+(z-2)^2}=1\) Square both side \((x-1)^{2}+(y+2)^{2}+(z-2)^{2}=1\) \(x^{2}-2 x+1+y^{2}+4 y+4+z^{2}-4 z+4=1\) Therefore locus of point \(\mathrm{P}\) is \(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-2 \mathrm{x}+4 \mathrm{y}-4 \mathrm{z}+8=0\)
CG PET-2010
Co-Ordinate system
88439
A variable plane is at a constant distance \(P\) from the origin \(O\) and meets the axes at \(A, B\) and \(C\). The locus of the centroid of the tetrahedron \(\mathrm{OABC}\) is
(B) : Let the plane meets then axes at \(\mathrm{A}(\mathrm{a}, 0,0)\), \(B(0, b, 0)\) and \(C(0,0, c)\) then, equation of plane is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\) It is given that plane is at a constant distance \(\mathrm{p}\) from the origin \((0,0,0)\) \(\therefore \quad \mathrm{p}=\left|\frac{0+0-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}}}\right|\) \(\Rightarrow \quad \frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}\) Let \((\alpha, \beta, \gamma)\) be the coordinates of centroid of the formed tetrahedron. Then, \(\alpha=\frac{a+0+0+0}{4} \Rightarrow a=4 \alpha\) \(\beta=\frac{0+b+0+0}{4} \Rightarrow b=4 \beta\) \(\gamma=\frac{0+0+c+0}{4} \Rightarrow c=4 \gamma\) On putting values of \(a b, c\) in Eq. (i) we get \(\frac{1}{\mathrm{p}^{2}}=\frac{1}{(4 \alpha)^{2}}+\frac{1}{(4 \beta)^{2}}+\frac{1}{(4 \gamma)^{2}}\) \(\Rightarrow \frac{1}{\mathrm{p}^{2}}=\frac{1}{16 \alpha^{2}}+\frac{1}{16 \beta^{2}}+\frac{1}{16 \gamma^{2}}\) \(\Rightarrow \frac{16}{p^{2}}=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}\) \(\therefore\) Locus of centroid of tetrahedron is \(\frac{16}{\mathrm{p}^{2}}=\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}+\frac{1}{\mathrm{z}^{2}}\)
CG PET-2015
Co-Ordinate system
88440
A line segment \(A B\) of length \(\lambda\) moves such that the points \(A\) and \(B\) remain on the periphery of a circle of radius \(\lambda\). Then the locus of the point, that divides the line segment \(A B\) in the ratio 2: 3 is a circle of radius
88441
A rod of length \(2 l\) slides with its ends on two perpendicular lines, then the locus of its midpoint is
1 \(\mathrm{x}^{2}+\mathrm{y}^{2}=l^{2}\)
2 \(\mathrm{x}^{2}-\mathrm{y}^{2}=l^{2}\)
3 \(2 \mathrm{x}^{2}+2 \mathrm{y}^{2}=l^{2}\)
4 \(2 x^{2}-2 y^{2}=l^{2}\)
Explanation:
(A): Let \(\mathrm{D}\) be the mid point on rod. Let \(\mathrm{x}=2 \alpha\) and \(\mathrm{y}=2 \beta\) \(\because x =2 \alpha \Rightarrow \alpha=\frac{x}{2}\) \(y =2 \beta \Rightarrow \beta=\frac{y}{2}\) In \(\triangle \mathrm{AOB}\), \(\mathrm{x}^{2}+\mathrm{y}^{2}=(2 l)^{2}\) \((2 \alpha)^{2}+(2 \beta)^{2}=4 l^{2}\) \(4 \alpha^{2}+4 \beta^{2}=4 l^{2}\) \(\alpha^{2}+\beta^{2}=l^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}=l^{2}\)
88438
The locus of the point, which moves such that its distance from \((1,-2,2)\) is unity, will be
1 \(x^{2}+y^{2}+z^{2}-2 x+4 y-4 z+8=0\)
2 \(x^{2}+y^{2}+z^{2}-2 x-4 y-4 z+8=0\)
3 \(x^{2}+y^{2}+z^{2}+2 x+4 y-4 z+8=0\)
4 \(x^{2}+y^{2}+z^{2}-2 x+4 y+4 z+8=0\)
Explanation:
(A) Let, The point in \(\mathrm{P}=(1,-2,2)\) So, \(\quad \mathrm{Q}=(\mathrm{x}, \mathrm{y}, \mathrm{z})\) Given \((\mathrm{PQ})=1\) \(\text { So, } \quad \sqrt{(x-1)^2+(y+2)^2+(z-2)^2}=1\) Square both side \((x-1)^{2}+(y+2)^{2}+(z-2)^{2}=1\) \(x^{2}-2 x+1+y^{2}+4 y+4+z^{2}-4 z+4=1\) Therefore locus of point \(\mathrm{P}\) is \(\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-2 \mathrm{x}+4 \mathrm{y}-4 \mathrm{z}+8=0\)
CG PET-2010
Co-Ordinate system
88439
A variable plane is at a constant distance \(P\) from the origin \(O\) and meets the axes at \(A, B\) and \(C\). The locus of the centroid of the tetrahedron \(\mathrm{OABC}\) is
(B) : Let the plane meets then axes at \(\mathrm{A}(\mathrm{a}, 0,0)\), \(B(0, b, 0)\) and \(C(0,0, c)\) then, equation of plane is \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\) It is given that plane is at a constant distance \(\mathrm{p}\) from the origin \((0,0,0)\) \(\therefore \quad \mathrm{p}=\left|\frac{0+0-1}{\sqrt{\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}}}\right|\) \(\Rightarrow \quad \frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}\) Let \((\alpha, \beta, \gamma)\) be the coordinates of centroid of the formed tetrahedron. Then, \(\alpha=\frac{a+0+0+0}{4} \Rightarrow a=4 \alpha\) \(\beta=\frac{0+b+0+0}{4} \Rightarrow b=4 \beta\) \(\gamma=\frac{0+0+c+0}{4} \Rightarrow c=4 \gamma\) On putting values of \(a b, c\) in Eq. (i) we get \(\frac{1}{\mathrm{p}^{2}}=\frac{1}{(4 \alpha)^{2}}+\frac{1}{(4 \beta)^{2}}+\frac{1}{(4 \gamma)^{2}}\) \(\Rightarrow \frac{1}{\mathrm{p}^{2}}=\frac{1}{16 \alpha^{2}}+\frac{1}{16 \beta^{2}}+\frac{1}{16 \gamma^{2}}\) \(\Rightarrow \frac{16}{p^{2}}=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}\) \(\therefore\) Locus of centroid of tetrahedron is \(\frac{16}{\mathrm{p}^{2}}=\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}+\frac{1}{\mathrm{z}^{2}}\)
CG PET-2015
Co-Ordinate system
88440
A line segment \(A B\) of length \(\lambda\) moves such that the points \(A\) and \(B\) remain on the periphery of a circle of radius \(\lambda\). Then the locus of the point, that divides the line segment \(A B\) in the ratio 2: 3 is a circle of radius
88441
A rod of length \(2 l\) slides with its ends on two perpendicular lines, then the locus of its midpoint is
1 \(\mathrm{x}^{2}+\mathrm{y}^{2}=l^{2}\)
2 \(\mathrm{x}^{2}-\mathrm{y}^{2}=l^{2}\)
3 \(2 \mathrm{x}^{2}+2 \mathrm{y}^{2}=l^{2}\)
4 \(2 x^{2}-2 y^{2}=l^{2}\)
Explanation:
(A): Let \(\mathrm{D}\) be the mid point on rod. Let \(\mathrm{x}=2 \alpha\) and \(\mathrm{y}=2 \beta\) \(\because x =2 \alpha \Rightarrow \alpha=\frac{x}{2}\) \(y =2 \beta \Rightarrow \beta=\frac{y}{2}\) In \(\triangle \mathrm{AOB}\), \(\mathrm{x}^{2}+\mathrm{y}^{2}=(2 l)^{2}\) \((2 \alpha)^{2}+(2 \beta)^{2}=4 l^{2}\) \(4 \alpha^{2}+4 \beta^{2}=4 l^{2}\) \(\alpha^{2}+\beta^{2}=l^{2}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}=l^{2}\)
