Equation of a Line, Sphere, and a Plane in Different Forms
« Previous Topic: Skew Lines and Coplanar Lines Next Topic: Angle Between Two Lines, Two Planes, a Line and a Plane »
Three Dimensional Geometry

121239 The distance of the point \((1,2,1)\) from the line \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}\) is

1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{2 \sqrt{5}}{3}\)
4 \(\frac{20}{3}\)
Karnataka CET-2019
Three Dimensional Geometry

121240 If the lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect, then \(k\) equals

1 \(\frac{3}{2}\)
2 \(\frac{9}{2}\)
3 \(-\frac{2}{9}\)
4 \(-\frac{3}{2}\)
MHT-CET-2020]
Three Dimensional Geometry

121241 What will be the distance of \((1,0,2)\) from the point of intersection of plane \(x-y+z=16\) and the line \(\left(\frac{x-2}{3}\right)=\left(\frac{y+1}{4}\right)=\left(\frac{z-2}{12}\right)\) ?

1 13 units
2 17 units
3 25 units
4 19 units
J and K CET-2017
Three Dimensional Geometry

121242 The equation of a plane containing the point \((1,-1,1)\) and parallel to the plane \(2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}=7\) is

1 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=-5\)
2 \(\bar{r} \cdot(2 \hat{i}-3 \hat{j}-4 \hat{k})=-1\)
3 \(\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3\)
4 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=5\)
MHT CET-2020
Three Dimensional Geometry

121244 The equation of line passing through the points \((3,4,-7)\) and \((6,-1,1)\) is

1 \(\frac{x-3}{-3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
2 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z-7}{8}\)
3 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
4 \(\frac{x-3}{3}=\frac{y-4}{5}=\frac{z+7}{8}\)
MHT CET-2020
NEET DIGITAL LIBRARY
Three Dimensional Geometry

121239 The distance of the point \((1,2,1)\) from the line \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}\) is

1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{2 \sqrt{5}}{3}\)
4 \(\frac{20}{3}\)
Karnataka CET-2019
Three Dimensional Geometry

121240 If the lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect, then \(k\) equals

1 \(\frac{3}{2}\)
2 \(\frac{9}{2}\)
3 \(-\frac{2}{9}\)
4 \(-\frac{3}{2}\)
MHT-CET-2020]
Three Dimensional Geometry

121241 What will be the distance of \((1,0,2)\) from the point of intersection of plane \(x-y+z=16\) and the line \(\left(\frac{x-2}{3}\right)=\left(\frac{y+1}{4}\right)=\left(\frac{z-2}{12}\right)\) ?

1 13 units
2 17 units
3 25 units
4 19 units
J and K CET-2017
Three Dimensional Geometry

121242 The equation of a plane containing the point \((1,-1,1)\) and parallel to the plane \(2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}=7\) is

1 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=-5\)
2 \(\bar{r} \cdot(2 \hat{i}-3 \hat{j}-4 \hat{k})=-1\)
3 \(\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3\)
4 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=5\)
MHT CET-2020
Three Dimensional Geometry

121244 The equation of line passing through the points \((3,4,-7)\) and \((6,-1,1)\) is

1 \(\frac{x-3}{-3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
2 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z-7}{8}\)
3 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
4 \(\frac{x-3}{3}=\frac{y-4}{5}=\frac{z+7}{8}\)
MHT CET-2020
Join With Us for More Updates
Three Dimensional Geometry

121239 The distance of the point \((1,2,1)\) from the line \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}\) is

1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{2 \sqrt{5}}{3}\)
4 \(\frac{20}{3}\)
Karnataka CET-2019
Three Dimensional Geometry

121240 If the lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect, then \(k\) equals

1 \(\frac{3}{2}\)
2 \(\frac{9}{2}\)
3 \(-\frac{2}{9}\)
4 \(-\frac{3}{2}\)
MHT-CET-2020]
Three Dimensional Geometry

121241 What will be the distance of \((1,0,2)\) from the point of intersection of plane \(x-y+z=16\) and the line \(\left(\frac{x-2}{3}\right)=\left(\frac{y+1}{4}\right)=\left(\frac{z-2}{12}\right)\) ?

1 13 units
2 17 units
3 25 units
4 19 units
J and K CET-2017
Three Dimensional Geometry

121242 The equation of a plane containing the point \((1,-1,1)\) and parallel to the plane \(2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}=7\) is

1 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=-5\)
2 \(\bar{r} \cdot(2 \hat{i}-3 \hat{j}-4 \hat{k})=-1\)
3 \(\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3\)
4 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=5\)
MHT CET-2020
Three Dimensional Geometry

121244 The equation of line passing through the points \((3,4,-7)\) and \((6,-1,1)\) is

1 \(\frac{x-3}{-3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
2 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z-7}{8}\)
3 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
4 \(\frac{x-3}{3}=\frac{y-4}{5}=\frac{z+7}{8}\)
MHT CET-2020
For More Updates join with Us
Three Dimensional Geometry

121239 The distance of the point \((1,2,1)\) from the line \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}\) is

1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{2 \sqrt{5}}{3}\)
4 \(\frac{20}{3}\)
Karnataka CET-2019
Three Dimensional Geometry

121240 If the lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect, then \(k\) equals

1 \(\frac{3}{2}\)
2 \(\frac{9}{2}\)
3 \(-\frac{2}{9}\)
4 \(-\frac{3}{2}\)
MHT-CET-2020]
Three Dimensional Geometry

121241 What will be the distance of \((1,0,2)\) from the point of intersection of plane \(x-y+z=16\) and the line \(\left(\frac{x-2}{3}\right)=\left(\frac{y+1}{4}\right)=\left(\frac{z-2}{12}\right)\) ?

1 13 units
2 17 units
3 25 units
4 19 units
J and K CET-2017
Three Dimensional Geometry

121242 The equation of a plane containing the point \((1,-1,1)\) and parallel to the plane \(2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}=7\) is

1 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=-5\)
2 \(\bar{r} \cdot(2 \hat{i}-3 \hat{j}-4 \hat{k})=-1\)
3 \(\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3\)
4 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=5\)
MHT CET-2020
Three Dimensional Geometry

121244 The equation of line passing through the points \((3,4,-7)\) and \((6,-1,1)\) is

1 \(\frac{x-3}{-3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
2 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z-7}{8}\)
3 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
4 \(\frac{x-3}{3}=\frac{y-4}{5}=\frac{z+7}{8}\)
MHT CET-2020
NEET DIGITAL LIBRARY
Three Dimensional Geometry

121239 The distance of the point \((1,2,1)\) from the line \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}\) is

1 \(\frac{2 \sqrt{3}}{5}\)
2 \(\frac{\sqrt{5}}{3}\)
3 \(\frac{2 \sqrt{5}}{3}\)
4 \(\frac{20}{3}\)
Karnataka CET-2019
Three Dimensional Geometry

121240 If the lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect, then \(k\) equals

1 \(\frac{3}{2}\)
2 \(\frac{9}{2}\)
3 \(-\frac{2}{9}\)
4 \(-\frac{3}{2}\)
MHT-CET-2020]
Three Dimensional Geometry

121241 What will be the distance of \((1,0,2)\) from the point of intersection of plane \(x-y+z=16\) and the line \(\left(\frac{x-2}{3}\right)=\left(\frac{y+1}{4}\right)=\left(\frac{z-2}{12}\right)\) ?

1 13 units
2 17 units
3 25 units
4 19 units
J and K CET-2017
Three Dimensional Geometry

121242 The equation of a plane containing the point \((1,-1,1)\) and parallel to the plane \(2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}=7\) is

1 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=-5\)
2 \(\bar{r} \cdot(2 \hat{i}-3 \hat{j}-4 \hat{k})=-1\)
3 \(\bar{r} \cdot(\hat{i}-\hat{j}+\hat{k})=3\)
4 \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=5\)
MHT CET-2020
Three Dimensional Geometry

121244 The equation of line passing through the points \((3,4,-7)\) and \((6,-1,1)\) is

1 \(\frac{x-3}{-3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
2 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z-7}{8}\)
3 \(\frac{x-3}{3}=\frac{y-4}{-5}=\frac{z+7}{8}\)
4 \(\frac{x-3}{3}=\frac{y-4}{5}=\frac{z+7}{8}\)
MHT CET-2020