121090
If the three vertices of a parallelogram \(A B C D\) are \(\mathrm{A}(1,0), \mathrm{B}(2,3), \mathrm{C}(3,2)\), what are the coordinates of the fourth point?
1 \((2,1)\)
2 \((2,-1)\)
3 \((-1,2)\)
4 \((-1,-2)\)
Explanation:
B As diagonal of parallelogram bisects each other. \(\therefore\) mid point of \(A C=\) mid point of \(B D\) \(\frac{3+1}{2}=\frac{2+x}{2} \text { and } \frac{0+2}{2}=\frac{3+y}{2}\) \(\Rightarrow x=2 \text { and } y=-1\)
SCRA-2012
Three Dimensional Geometry
121091
The foot of perpendicular from the point \((3,4,5)\) to the plane \(x+y+z=9\) is
1 \((2,3,4)\)
2 \((3,5,-2)\)
3 \((3,5,2)\)
4 \((3,2,4)\)
Explanation:
A Given, Point \(\mathrm{P}=(3,4,5)\) Plane, \(\quad x+y+z=9\) Equation of line PQ perpendicular to given plane is- \(\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{1}=\lambda\) \(\Rightarrow \quad \mathrm{x}=\lambda+3\) \(\mathrm{y}=\lambda+4\) and \(\quad z=\lambda+5\) Now, point \(\mathrm{Q}\) lies on the plane. So, \(\quad \lambda+3+\lambda+4+\lambda+5=9\) \(\Rightarrow \quad 3 \lambda+12=9\) \(\Rightarrow \quad \lambda=-1\) So, the coordinates of the foot of the perpendicular be \((2,3,4)\) \(\therefore(\mathrm{x}, \mathrm{y}, \mathrm{z})=(2,3,4)\)
CG PET- 2012
Three Dimensional Geometry
121093
Let \(P\) be the plane, passing through the point \((1,-1,-5)\) and perpendicular to the line joining the points \((4,1,-3)\) and \((2,4,3)\). Then the distance of \(P\) from the point \((3,-2,2)\) is
1 6
2 4
3 5
4 7
Explanation:
C Let the given point be \(A=(1,-1,5)\)
\(B=(4,1,-3)\)
\(C=(2,4,3)\)
Then, equation of the plane \(2(\mathrm{x}-1)-3(\mathrm{y}+1)-6(\mathrm{z}+5)=0\).
or \(2 x-3 y-6 z=35\)
Distance, \(d=\left \vert\frac{2 \times 3-3 \times(-2)-6 \times 2-35}{\sqrt{4+9+36}}\right \vert\)
\(=\left \vert\frac{6+6-12-35}{\sqrt{4+9+36}}\right \vert=\frac{35}{7}\)
\(d =5\)
JEE Main-31.01.2023
Three Dimensional Geometry
121094
The foot of perpendicular from the origin \(O\) to a plane \(P\) which meets the co-ordinate axes at the point \(A, B, C\) is \((2, a, 4), a \in N\). If the volume of the tetrahedron OAB is 144 unit \(^3\), then which of the following points is NOT on \(P\) ?
1 \((0,4,4)\)
2 \((3,0,4)\)
3 \((0,6,3)\)
4 \((2,2,4)\)
Explanation:
B Given, Equation of plane is - \(2(x-2)+a(y-a)+4(z-4)=0 .\) \(2 x+a y+4 z=20+a^2\) Again - \(\frac{x}{\left(\frac{20+\mathrm{a}^2}{2}\right)}+\frac{y}{\left(\frac{20+\mathrm{a}^2}{\mathrm{a}}\right)}+\frac{\mathrm{z}}{\left(\frac{20+\mathrm{a}^2}{4}\right)}=1\) Volume of tetrahedron \(\mathrm{OABC}=\frac{1}{6} \mathrm{abc}\) \(\Rightarrow \frac{1}{6}\left(\frac{20+\mathrm{a}^2}{2}\right)\left(\frac{20+\mathrm{a}^2}{\mathrm{a}}\right) \cdot\left(\frac{20+\mathrm{a}^2}{4}\right)=144\) \(\Rightarrow\left(20+\mathrm{a}^2\right)^3=\mathrm{a} \times 48 \times 144\) \(\Rightarrow\left(20+\mathrm{a}^2\right)^3=12^3(4 \mathrm{a})\) This equation satisfies, if \(a=2\) So, plane \(\mathrm{P}\) is \(2 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=24\) \(\mathrm{x}+\mathrm{y}+2 \mathrm{z}=12\)Point \((3,0,4)\) not satisfies the plane.
121090
If the three vertices of a parallelogram \(A B C D\) are \(\mathrm{A}(1,0), \mathrm{B}(2,3), \mathrm{C}(3,2)\), what are the coordinates of the fourth point?
1 \((2,1)\)
2 \((2,-1)\)
3 \((-1,2)\)
4 \((-1,-2)\)
Explanation:
B As diagonal of parallelogram bisects each other. \(\therefore\) mid point of \(A C=\) mid point of \(B D\) \(\frac{3+1}{2}=\frac{2+x}{2} \text { and } \frac{0+2}{2}=\frac{3+y}{2}\) \(\Rightarrow x=2 \text { and } y=-1\)
SCRA-2012
Three Dimensional Geometry
121091
The foot of perpendicular from the point \((3,4,5)\) to the plane \(x+y+z=9\) is
1 \((2,3,4)\)
2 \((3,5,-2)\)
3 \((3,5,2)\)
4 \((3,2,4)\)
Explanation:
A Given, Point \(\mathrm{P}=(3,4,5)\) Plane, \(\quad x+y+z=9\) Equation of line PQ perpendicular to given plane is- \(\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{1}=\lambda\) \(\Rightarrow \quad \mathrm{x}=\lambda+3\) \(\mathrm{y}=\lambda+4\) and \(\quad z=\lambda+5\) Now, point \(\mathrm{Q}\) lies on the plane. So, \(\quad \lambda+3+\lambda+4+\lambda+5=9\) \(\Rightarrow \quad 3 \lambda+12=9\) \(\Rightarrow \quad \lambda=-1\) So, the coordinates of the foot of the perpendicular be \((2,3,4)\) \(\therefore(\mathrm{x}, \mathrm{y}, \mathrm{z})=(2,3,4)\)
CG PET- 2012
Three Dimensional Geometry
121093
Let \(P\) be the plane, passing through the point \((1,-1,-5)\) and perpendicular to the line joining the points \((4,1,-3)\) and \((2,4,3)\). Then the distance of \(P\) from the point \((3,-2,2)\) is
1 6
2 4
3 5
4 7
Explanation:
C Let the given point be \(A=(1,-1,5)\)
\(B=(4,1,-3)\)
\(C=(2,4,3)\)
Then, equation of the plane \(2(\mathrm{x}-1)-3(\mathrm{y}+1)-6(\mathrm{z}+5)=0\).
or \(2 x-3 y-6 z=35\)
Distance, \(d=\left \vert\frac{2 \times 3-3 \times(-2)-6 \times 2-35}{\sqrt{4+9+36}}\right \vert\)
\(=\left \vert\frac{6+6-12-35}{\sqrt{4+9+36}}\right \vert=\frac{35}{7}\)
\(d =5\)
JEE Main-31.01.2023
Three Dimensional Geometry
121094
The foot of perpendicular from the origin \(O\) to a plane \(P\) which meets the co-ordinate axes at the point \(A, B, C\) is \((2, a, 4), a \in N\). If the volume of the tetrahedron OAB is 144 unit \(^3\), then which of the following points is NOT on \(P\) ?
1 \((0,4,4)\)
2 \((3,0,4)\)
3 \((0,6,3)\)
4 \((2,2,4)\)
Explanation:
B Given, Equation of plane is - \(2(x-2)+a(y-a)+4(z-4)=0 .\) \(2 x+a y+4 z=20+a^2\) Again - \(\frac{x}{\left(\frac{20+\mathrm{a}^2}{2}\right)}+\frac{y}{\left(\frac{20+\mathrm{a}^2}{\mathrm{a}}\right)}+\frac{\mathrm{z}}{\left(\frac{20+\mathrm{a}^2}{4}\right)}=1\) Volume of tetrahedron \(\mathrm{OABC}=\frac{1}{6} \mathrm{abc}\) \(\Rightarrow \frac{1}{6}\left(\frac{20+\mathrm{a}^2}{2}\right)\left(\frac{20+\mathrm{a}^2}{\mathrm{a}}\right) \cdot\left(\frac{20+\mathrm{a}^2}{4}\right)=144\) \(\Rightarrow\left(20+\mathrm{a}^2\right)^3=\mathrm{a} \times 48 \times 144\) \(\Rightarrow\left(20+\mathrm{a}^2\right)^3=12^3(4 \mathrm{a})\) This equation satisfies, if \(a=2\) So, plane \(\mathrm{P}\) is \(2 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=24\) \(\mathrm{x}+\mathrm{y}+2 \mathrm{z}=12\)Point \((3,0,4)\) not satisfies the plane.
121090
If the three vertices of a parallelogram \(A B C D\) are \(\mathrm{A}(1,0), \mathrm{B}(2,3), \mathrm{C}(3,2)\), what are the coordinates of the fourth point?
1 \((2,1)\)
2 \((2,-1)\)
3 \((-1,2)\)
4 \((-1,-2)\)
Explanation:
B As diagonal of parallelogram bisects each other. \(\therefore\) mid point of \(A C=\) mid point of \(B D\) \(\frac{3+1}{2}=\frac{2+x}{2} \text { and } \frac{0+2}{2}=\frac{3+y}{2}\) \(\Rightarrow x=2 \text { and } y=-1\)
SCRA-2012
Three Dimensional Geometry
121091
The foot of perpendicular from the point \((3,4,5)\) to the plane \(x+y+z=9\) is
1 \((2,3,4)\)
2 \((3,5,-2)\)
3 \((3,5,2)\)
4 \((3,2,4)\)
Explanation:
A Given, Point \(\mathrm{P}=(3,4,5)\) Plane, \(\quad x+y+z=9\) Equation of line PQ perpendicular to given plane is- \(\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{1}=\lambda\) \(\Rightarrow \quad \mathrm{x}=\lambda+3\) \(\mathrm{y}=\lambda+4\) and \(\quad z=\lambda+5\) Now, point \(\mathrm{Q}\) lies on the plane. So, \(\quad \lambda+3+\lambda+4+\lambda+5=9\) \(\Rightarrow \quad 3 \lambda+12=9\) \(\Rightarrow \quad \lambda=-1\) So, the coordinates of the foot of the perpendicular be \((2,3,4)\) \(\therefore(\mathrm{x}, \mathrm{y}, \mathrm{z})=(2,3,4)\)
CG PET- 2012
Three Dimensional Geometry
121093
Let \(P\) be the plane, passing through the point \((1,-1,-5)\) and perpendicular to the line joining the points \((4,1,-3)\) and \((2,4,3)\). Then the distance of \(P\) from the point \((3,-2,2)\) is
1 6
2 4
3 5
4 7
Explanation:
C Let the given point be \(A=(1,-1,5)\)
\(B=(4,1,-3)\)
\(C=(2,4,3)\)
Then, equation of the plane \(2(\mathrm{x}-1)-3(\mathrm{y}+1)-6(\mathrm{z}+5)=0\).
or \(2 x-3 y-6 z=35\)
Distance, \(d=\left \vert\frac{2 \times 3-3 \times(-2)-6 \times 2-35}{\sqrt{4+9+36}}\right \vert\)
\(=\left \vert\frac{6+6-12-35}{\sqrt{4+9+36}}\right \vert=\frac{35}{7}\)
\(d =5\)
JEE Main-31.01.2023
Three Dimensional Geometry
121094
The foot of perpendicular from the origin \(O\) to a plane \(P\) which meets the co-ordinate axes at the point \(A, B, C\) is \((2, a, 4), a \in N\). If the volume of the tetrahedron OAB is 144 unit \(^3\), then which of the following points is NOT on \(P\) ?
1 \((0,4,4)\)
2 \((3,0,4)\)
3 \((0,6,3)\)
4 \((2,2,4)\)
Explanation:
B Given, Equation of plane is - \(2(x-2)+a(y-a)+4(z-4)=0 .\) \(2 x+a y+4 z=20+a^2\) Again - \(\frac{x}{\left(\frac{20+\mathrm{a}^2}{2}\right)}+\frac{y}{\left(\frac{20+\mathrm{a}^2}{\mathrm{a}}\right)}+\frac{\mathrm{z}}{\left(\frac{20+\mathrm{a}^2}{4}\right)}=1\) Volume of tetrahedron \(\mathrm{OABC}=\frac{1}{6} \mathrm{abc}\) \(\Rightarrow \frac{1}{6}\left(\frac{20+\mathrm{a}^2}{2}\right)\left(\frac{20+\mathrm{a}^2}{\mathrm{a}}\right) \cdot\left(\frac{20+\mathrm{a}^2}{4}\right)=144\) \(\Rightarrow\left(20+\mathrm{a}^2\right)^3=\mathrm{a} \times 48 \times 144\) \(\Rightarrow\left(20+\mathrm{a}^2\right)^3=12^3(4 \mathrm{a})\) This equation satisfies, if \(a=2\) So, plane \(\mathrm{P}\) is \(2 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=24\) \(\mathrm{x}+\mathrm{y}+2 \mathrm{z}=12\)Point \((3,0,4)\) not satisfies the plane.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Three Dimensional Geometry
121090
If the three vertices of a parallelogram \(A B C D\) are \(\mathrm{A}(1,0), \mathrm{B}(2,3), \mathrm{C}(3,2)\), what are the coordinates of the fourth point?
1 \((2,1)\)
2 \((2,-1)\)
3 \((-1,2)\)
4 \((-1,-2)\)
Explanation:
B As diagonal of parallelogram bisects each other. \(\therefore\) mid point of \(A C=\) mid point of \(B D\) \(\frac{3+1}{2}=\frac{2+x}{2} \text { and } \frac{0+2}{2}=\frac{3+y}{2}\) \(\Rightarrow x=2 \text { and } y=-1\)
SCRA-2012
Three Dimensional Geometry
121091
The foot of perpendicular from the point \((3,4,5)\) to the plane \(x+y+z=9\) is
1 \((2,3,4)\)
2 \((3,5,-2)\)
3 \((3,5,2)\)
4 \((3,2,4)\)
Explanation:
A Given, Point \(\mathrm{P}=(3,4,5)\) Plane, \(\quad x+y+z=9\) Equation of line PQ perpendicular to given plane is- \(\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{1}=\lambda\) \(\Rightarrow \quad \mathrm{x}=\lambda+3\) \(\mathrm{y}=\lambda+4\) and \(\quad z=\lambda+5\) Now, point \(\mathrm{Q}\) lies on the plane. So, \(\quad \lambda+3+\lambda+4+\lambda+5=9\) \(\Rightarrow \quad 3 \lambda+12=9\) \(\Rightarrow \quad \lambda=-1\) So, the coordinates of the foot of the perpendicular be \((2,3,4)\) \(\therefore(\mathrm{x}, \mathrm{y}, \mathrm{z})=(2,3,4)\)
CG PET- 2012
Three Dimensional Geometry
121093
Let \(P\) be the plane, passing through the point \((1,-1,-5)\) and perpendicular to the line joining the points \((4,1,-3)\) and \((2,4,3)\). Then the distance of \(P\) from the point \((3,-2,2)\) is
1 6
2 4
3 5
4 7
Explanation:
C Let the given point be \(A=(1,-1,5)\)
\(B=(4,1,-3)\)
\(C=(2,4,3)\)
Then, equation of the plane \(2(\mathrm{x}-1)-3(\mathrm{y}+1)-6(\mathrm{z}+5)=0\).
or \(2 x-3 y-6 z=35\)
Distance, \(d=\left \vert\frac{2 \times 3-3 \times(-2)-6 \times 2-35}{\sqrt{4+9+36}}\right \vert\)
\(=\left \vert\frac{6+6-12-35}{\sqrt{4+9+36}}\right \vert=\frac{35}{7}\)
\(d =5\)
JEE Main-31.01.2023
Three Dimensional Geometry
121094
The foot of perpendicular from the origin \(O\) to a plane \(P\) which meets the co-ordinate axes at the point \(A, B, C\) is \((2, a, 4), a \in N\). If the volume of the tetrahedron OAB is 144 unit \(^3\), then which of the following points is NOT on \(P\) ?
1 \((0,4,4)\)
2 \((3,0,4)\)
3 \((0,6,3)\)
4 \((2,2,4)\)
Explanation:
B Given, Equation of plane is - \(2(x-2)+a(y-a)+4(z-4)=0 .\) \(2 x+a y+4 z=20+a^2\) Again - \(\frac{x}{\left(\frac{20+\mathrm{a}^2}{2}\right)}+\frac{y}{\left(\frac{20+\mathrm{a}^2}{\mathrm{a}}\right)}+\frac{\mathrm{z}}{\left(\frac{20+\mathrm{a}^2}{4}\right)}=1\) Volume of tetrahedron \(\mathrm{OABC}=\frac{1}{6} \mathrm{abc}\) \(\Rightarrow \frac{1}{6}\left(\frac{20+\mathrm{a}^2}{2}\right)\left(\frac{20+\mathrm{a}^2}{\mathrm{a}}\right) \cdot\left(\frac{20+\mathrm{a}^2}{4}\right)=144\) \(\Rightarrow\left(20+\mathrm{a}^2\right)^3=\mathrm{a} \times 48 \times 144\) \(\Rightarrow\left(20+\mathrm{a}^2\right)^3=12^3(4 \mathrm{a})\) This equation satisfies, if \(a=2\) So, plane \(\mathrm{P}\) is \(2 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}=24\) \(\mathrm{x}+\mathrm{y}+2 \mathrm{z}=12\)Point \((3,0,4)\) not satisfies the plane.
