Miscellaneous 3-D Problems
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Three Dimensional Geometry

121400 Equation of two straight lines are \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-4}{5}=\frac{y-1}{2}=z\).
Then

1 The lines are non-coplanar
2 The lines are parallel and distinct
3 The lines intersect in unique point
4 The lines are coincident
BITSAT-2019
Three Dimensional Geometry

121401 The lines \(\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}\) and \(\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}\) intersect at the point

1 \((11,-4,5)\)
2 \((-11,-4,5)\)
3 \((11,4,-5)\)
4 \((-11,-4,-5)\)
MHT CET-2013
Three Dimensional Geometry

121402 The value of \(k\) such that the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane \(2 x-4 y+z=7\) is

1 -7
2 4
3 -4
4 7
Karnataka CET-2018
Three Dimensional Geometry

121403 The locus represented by \(x y+y z=0\) is

1 a pair of perpendicular lines
2 a pair of parallel lines
3 a pair of parallel planes
4 a pair of perpendicular planes
Karnataka CET-2018
Three Dimensional Geometry

121405 Equation of the plane perpendicular to the line \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and passing through the point \((2,3\), 4) is

1 \(2 x+3 y+z=17\)
2 \(x+2 y+3 z=9\)
3 \(3 x+2 y+z=16\)
4 \(x+2 y+3 z=20\)
Karnataka CET-2014
NEET DIGITAL LIBRARY
Three Dimensional Geometry

121400 Equation of two straight lines are \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-4}{5}=\frac{y-1}{2}=z\).
Then

1 The lines are non-coplanar
2 The lines are parallel and distinct
3 The lines intersect in unique point
4 The lines are coincident
BITSAT-2019
Three Dimensional Geometry

121401 The lines \(\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}\) and \(\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}\) intersect at the point

1 \((11,-4,5)\)
2 \((-11,-4,5)\)
3 \((11,4,-5)\)
4 \((-11,-4,-5)\)
MHT CET-2013
Three Dimensional Geometry

121402 The value of \(k\) such that the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane \(2 x-4 y+z=7\) is

1 -7
2 4
3 -4
4 7
Karnataka CET-2018
Three Dimensional Geometry

121403 The locus represented by \(x y+y z=0\) is

1 a pair of perpendicular lines
2 a pair of parallel lines
3 a pair of parallel planes
4 a pair of perpendicular planes
Karnataka CET-2018
Three Dimensional Geometry

121405 Equation of the plane perpendicular to the line \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and passing through the point \((2,3\), 4) is

1 \(2 x+3 y+z=17\)
2 \(x+2 y+3 z=9\)
3 \(3 x+2 y+z=16\)
4 \(x+2 y+3 z=20\)
Karnataka CET-2014
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Three Dimensional Geometry

121400 Equation of two straight lines are \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-4}{5}=\frac{y-1}{2}=z\).
Then

1 The lines are non-coplanar
2 The lines are parallel and distinct
3 The lines intersect in unique point
4 The lines are coincident
BITSAT-2019
Three Dimensional Geometry

121401 The lines \(\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}\) and \(\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}\) intersect at the point

1 \((11,-4,5)\)
2 \((-11,-4,5)\)
3 \((11,4,-5)\)
4 \((-11,-4,-5)\)
MHT CET-2013
Three Dimensional Geometry

121402 The value of \(k\) such that the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane \(2 x-4 y+z=7\) is

1 -7
2 4
3 -4
4 7
Karnataka CET-2018
Three Dimensional Geometry

121403 The locus represented by \(x y+y z=0\) is

1 a pair of perpendicular lines
2 a pair of parallel lines
3 a pair of parallel planes
4 a pair of perpendicular planes
Karnataka CET-2018
Three Dimensional Geometry

121405 Equation of the plane perpendicular to the line \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and passing through the point \((2,3\), 4) is

1 \(2 x+3 y+z=17\)
2 \(x+2 y+3 z=9\)
3 \(3 x+2 y+z=16\)
4 \(x+2 y+3 z=20\)
Karnataka CET-2014
For More Updates join with Us
Three Dimensional Geometry

121400 Equation of two straight lines are \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-4}{5}=\frac{y-1}{2}=z\).
Then

1 The lines are non-coplanar
2 The lines are parallel and distinct
3 The lines intersect in unique point
4 The lines are coincident
BITSAT-2019
Three Dimensional Geometry

121401 The lines \(\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}\) and \(\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}\) intersect at the point

1 \((11,-4,5)\)
2 \((-11,-4,5)\)
3 \((11,4,-5)\)
4 \((-11,-4,-5)\)
MHT CET-2013
Three Dimensional Geometry

121402 The value of \(k\) such that the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane \(2 x-4 y+z=7\) is

1 -7
2 4
3 -4
4 7
Karnataka CET-2018
Three Dimensional Geometry

121403 The locus represented by \(x y+y z=0\) is

1 a pair of perpendicular lines
2 a pair of parallel lines
3 a pair of parallel planes
4 a pair of perpendicular planes
Karnataka CET-2018
Three Dimensional Geometry

121405 Equation of the plane perpendicular to the line \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and passing through the point \((2,3\), 4) is

1 \(2 x+3 y+z=17\)
2 \(x+2 y+3 z=9\)
3 \(3 x+2 y+z=16\)
4 \(x+2 y+3 z=20\)
Karnataka CET-2014
NEET DIGITAL LIBRARY
Three Dimensional Geometry

121400 Equation of two straight lines are \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-4}{5}=\frac{y-1}{2}=z\).
Then

1 The lines are non-coplanar
2 The lines are parallel and distinct
3 The lines intersect in unique point
4 The lines are coincident
BITSAT-2019
Three Dimensional Geometry

121401 The lines \(\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}\) and \(\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}\) intersect at the point

1 \((11,-4,5)\)
2 \((-11,-4,5)\)
3 \((11,4,-5)\)
4 \((-11,-4,-5)\)
MHT CET-2013
Three Dimensional Geometry

121402 The value of \(k\) such that the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane \(2 x-4 y+z=7\) is

1 -7
2 4
3 -4
4 7
Karnataka CET-2018
Three Dimensional Geometry

121403 The locus represented by \(x y+y z=0\) is

1 a pair of perpendicular lines
2 a pair of parallel lines
3 a pair of parallel planes
4 a pair of perpendicular planes
Karnataka CET-2018
Three Dimensional Geometry

121405 Equation of the plane perpendicular to the line \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and passing through the point \((2,3\), 4) is

1 \(2 x+3 y+z=17\)
2 \(x+2 y+3 z=9\)
3 \(3 x+2 y+z=16\)
4 \(x+2 y+3 z=20\)
Karnataka CET-2014