Miscellaneous 3-D Problems
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Three Dimensional Geometry

121440 Let the plane \(\pi\) pass through the point \((1,0,1)\) and perpendicular to the planes \(2 x+3 y-z=2\) and \(x-y+2 z=1\). Let the equation of the plane passing through the point \((11,7,5)\) and parallel to the plane \(\pi\) be \(a x+b y-z+d=0\). Then \(\frac{\mathbf{a}}{\mathbf{b}}+\frac{\mathbf{b}}{\mathbf{d}}=\)

1 3
2 0
3 2
4 -2
AP EAMCET-05.07.2022
Three Dimensional Geometry

121441 For \(a \neq 0\), if the sum of the distances of a point from the points \((a, 0,0)\) and \((-a, 0,0)\) constant \(2 k\), then the locus of that point is

1 \(\mathrm{x}^2+\mathrm{k}^2\left(\mathrm{y}^2+\mathrm{z}^2\right)=\mathrm{k}^2\)
2 \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=\frac{1}{\mathrm{k}^2+1}\)
3 \(\frac{\mathrm{x}^2}{\mathrm{k}^2}+\frac{\mathrm{y}^2+\mathrm{z}^2}{\mathrm{k}^2-\mathrm{a}^2}=1\)
4 \(\frac{\mathrm{x}^2}{\mathrm{k}^2-\mathrm{a}^2}+\frac{\mathrm{y}^2+\mathrm{z}^2}{\mathrm{k}^2}=1\)
AP EAMCET-24.04.2018
Three Dimensional Geometry

121417 The perpendicular distance of the point \(P(6,7\), 8) from XY-plane is

1 7
2 6
3 8
4 5
Karnataka CET-2017
Three Dimensional Geometry

121428 If the plane \(7 x+11 y+13 z=3003\) meets the coordinate axes in \(A, B, C\), then the centroid of the \(\triangle \mathrm{ABC}\) is

1 \((143,91,77)\)
2 \((143,77,91)\)
3 \((91,143,77)\)
4 \((143,66,91)\)
AP EAMCET-2002
NEET DIGITAL LIBRARY
Three Dimensional Geometry

121440 Let the plane \(\pi\) pass through the point \((1,0,1)\) and perpendicular to the planes \(2 x+3 y-z=2\) and \(x-y+2 z=1\). Let the equation of the plane passing through the point \((11,7,5)\) and parallel to the plane \(\pi\) be \(a x+b y-z+d=0\). Then \(\frac{\mathbf{a}}{\mathbf{b}}+\frac{\mathbf{b}}{\mathbf{d}}=\)

1 3
2 0
3 2
4 -2
AP EAMCET-05.07.2022
Three Dimensional Geometry

121441 For \(a \neq 0\), if the sum of the distances of a point from the points \((a, 0,0)\) and \((-a, 0,0)\) constant \(2 k\), then the locus of that point is

1 \(\mathrm{x}^2+\mathrm{k}^2\left(\mathrm{y}^2+\mathrm{z}^2\right)=\mathrm{k}^2\)
2 \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=\frac{1}{\mathrm{k}^2+1}\)
3 \(\frac{\mathrm{x}^2}{\mathrm{k}^2}+\frac{\mathrm{y}^2+\mathrm{z}^2}{\mathrm{k}^2-\mathrm{a}^2}=1\)
4 \(\frac{\mathrm{x}^2}{\mathrm{k}^2-\mathrm{a}^2}+\frac{\mathrm{y}^2+\mathrm{z}^2}{\mathrm{k}^2}=1\)
AP EAMCET-24.04.2018
Three Dimensional Geometry

121417 The perpendicular distance of the point \(P(6,7\), 8) from XY-plane is

1 7
2 6
3 8
4 5
Karnataka CET-2017
Three Dimensional Geometry

121428 If the plane \(7 x+11 y+13 z=3003\) meets the coordinate axes in \(A, B, C\), then the centroid of the \(\triangle \mathrm{ABC}\) is

1 \((143,91,77)\)
2 \((143,77,91)\)
3 \((91,143,77)\)
4 \((143,66,91)\)
AP EAMCET-2002
Join With Us for More Updates
Three Dimensional Geometry

121440 Let the plane \(\pi\) pass through the point \((1,0,1)\) and perpendicular to the planes \(2 x+3 y-z=2\) and \(x-y+2 z=1\). Let the equation of the plane passing through the point \((11,7,5)\) and parallel to the plane \(\pi\) be \(a x+b y-z+d=0\). Then \(\frac{\mathbf{a}}{\mathbf{b}}+\frac{\mathbf{b}}{\mathbf{d}}=\)

1 3
2 0
3 2
4 -2
AP EAMCET-05.07.2022
Three Dimensional Geometry

121441 For \(a \neq 0\), if the sum of the distances of a point from the points \((a, 0,0)\) and \((-a, 0,0)\) constant \(2 k\), then the locus of that point is

1 \(\mathrm{x}^2+\mathrm{k}^2\left(\mathrm{y}^2+\mathrm{z}^2\right)=\mathrm{k}^2\)
2 \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=\frac{1}{\mathrm{k}^2+1}\)
3 \(\frac{\mathrm{x}^2}{\mathrm{k}^2}+\frac{\mathrm{y}^2+\mathrm{z}^2}{\mathrm{k}^2-\mathrm{a}^2}=1\)
4 \(\frac{\mathrm{x}^2}{\mathrm{k}^2-\mathrm{a}^2}+\frac{\mathrm{y}^2+\mathrm{z}^2}{\mathrm{k}^2}=1\)
AP EAMCET-24.04.2018
Three Dimensional Geometry

121417 The perpendicular distance of the point \(P(6,7\), 8) from XY-plane is

1 7
2 6
3 8
4 5
Karnataka CET-2017
Three Dimensional Geometry

121428 If the plane \(7 x+11 y+13 z=3003\) meets the coordinate axes in \(A, B, C\), then the centroid of the \(\triangle \mathrm{ABC}\) is

1 \((143,91,77)\)
2 \((143,77,91)\)
3 \((91,143,77)\)
4 \((143,66,91)\)
AP EAMCET-2002
For More Updates join with Us
Three Dimensional Geometry

121440 Let the plane \(\pi\) pass through the point \((1,0,1)\) and perpendicular to the planes \(2 x+3 y-z=2\) and \(x-y+2 z=1\). Let the equation of the plane passing through the point \((11,7,5)\) and parallel to the plane \(\pi\) be \(a x+b y-z+d=0\). Then \(\frac{\mathbf{a}}{\mathbf{b}}+\frac{\mathbf{b}}{\mathbf{d}}=\)

1 3
2 0
3 2
4 -2
AP EAMCET-05.07.2022
Three Dimensional Geometry

121441 For \(a \neq 0\), if the sum of the distances of a point from the points \((a, 0,0)\) and \((-a, 0,0)\) constant \(2 k\), then the locus of that point is

1 \(\mathrm{x}^2+\mathrm{k}^2\left(\mathrm{y}^2+\mathrm{z}^2\right)=\mathrm{k}^2\)
2 \(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=\frac{1}{\mathrm{k}^2+1}\)
3 \(\frac{\mathrm{x}^2}{\mathrm{k}^2}+\frac{\mathrm{y}^2+\mathrm{z}^2}{\mathrm{k}^2-\mathrm{a}^2}=1\)
4 \(\frac{\mathrm{x}^2}{\mathrm{k}^2-\mathrm{a}^2}+\frac{\mathrm{y}^2+\mathrm{z}^2}{\mathrm{k}^2}=1\)
AP EAMCET-24.04.2018
Three Dimensional Geometry

121417 The perpendicular distance of the point \(P(6,7\), 8) from XY-plane is

1 7
2 6
3 8
4 5
Karnataka CET-2017
Three Dimensional Geometry

121428 If the plane \(7 x+11 y+13 z=3003\) meets the coordinate axes in \(A, B, C\), then the centroid of the \(\triangle \mathrm{ABC}\) is

1 \((143,91,77)\)
2 \((143,77,91)\)
3 \((91,143,77)\)
4 \((143,66,91)\)
AP EAMCET-2002
NEET DIGITAL LIBRARY