Cartesian and Vector Equations of Line and Plane
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Three Dimensional Geometry

121200 The equation of the line joining the points \((-3,4,11)\) and \((1,-2,7)\) is

1 \(\frac{x+3}{2}=\frac{y-4}{3}=\frac{z-11}{4}\)
2 \(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\)
3 \(\frac{x+3}{-2}=\frac{y+4}{3}=\frac{z+11}{4}\)
4 \(\frac{x+3}{2}=\frac{y+4}{-3}=\frac{z+11}{2}\)
Karnataka CET-2021
Three Dimensional Geometry

121201 If the planes \(\overrightarrow{\mathbf{r}} \cdot(2 \hat{\mathbf{i}}-\lambda \hat{\mathbf{j}}+\hat{\mathbf{k}})=3\) and \(\overrightarrow{\mathbf{r}} \cdot(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mu \hat{\mathbf{k}})=5\) are parallel, then what is \(\lambda\) equal to?

1 \(1 / 2\)
2 1
3 \(-1 / 2\)
4 -1
SCRA-2012
Three Dimensional Geometry

121203 The Cartesian equation of the line passing through the point \((-1,3,-2)\) and perpendicular to the lines \(\quad \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}\) is

1 \(\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}\)
2 \(\frac{x-1}{-2}=\frac{y+3}{-7}=\frac{z-2}{-4}\)
3 \(\frac{x+1}{2}=\frac{y+3}{7}=\frac{z+2}{4}\)
4 \(\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry

121204 The equation of the plane, which bisects the line joining the points \((1,2,3)\) and \((3,4,5)\) at right angles is

1 \(x+y+z=0\)
2 \(x+y-z=0\)
3 \(x+y+z=9\)
4 \(\mathrm{x}+\mathrm{y}-\mathrm{z}+9=0\)
WB JEE-2019
NEET DIGITAL LIBRARY
Three Dimensional Geometry

121200 The equation of the line joining the points \((-3,4,11)\) and \((1,-2,7)\) is

1 \(\frac{x+3}{2}=\frac{y-4}{3}=\frac{z-11}{4}\)
2 \(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\)
3 \(\frac{x+3}{-2}=\frac{y+4}{3}=\frac{z+11}{4}\)
4 \(\frac{x+3}{2}=\frac{y+4}{-3}=\frac{z+11}{2}\)
Karnataka CET-2021
Three Dimensional Geometry

121201 If the planes \(\overrightarrow{\mathbf{r}} \cdot(2 \hat{\mathbf{i}}-\lambda \hat{\mathbf{j}}+\hat{\mathbf{k}})=3\) and \(\overrightarrow{\mathbf{r}} \cdot(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mu \hat{\mathbf{k}})=5\) are parallel, then what is \(\lambda\) equal to?

1 \(1 / 2\)
2 1
3 \(-1 / 2\)
4 -1
SCRA-2012
Three Dimensional Geometry

121203 The Cartesian equation of the line passing through the point \((-1,3,-2)\) and perpendicular to the lines \(\quad \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}\) is

1 \(\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}\)
2 \(\frac{x-1}{-2}=\frac{y+3}{-7}=\frac{z-2}{-4}\)
3 \(\frac{x+1}{2}=\frac{y+3}{7}=\frac{z+2}{4}\)
4 \(\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry

121204 The equation of the plane, which bisects the line joining the points \((1,2,3)\) and \((3,4,5)\) at right angles is

1 \(x+y+z=0\)
2 \(x+y-z=0\)
3 \(x+y+z=9\)
4 \(\mathrm{x}+\mathrm{y}-\mathrm{z}+9=0\)
WB JEE-2019
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Three Dimensional Geometry

121200 The equation of the line joining the points \((-3,4,11)\) and \((1,-2,7)\) is

1 \(\frac{x+3}{2}=\frac{y-4}{3}=\frac{z-11}{4}\)
2 \(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\)
3 \(\frac{x+3}{-2}=\frac{y+4}{3}=\frac{z+11}{4}\)
4 \(\frac{x+3}{2}=\frac{y+4}{-3}=\frac{z+11}{2}\)
Karnataka CET-2021
Three Dimensional Geometry

121201 If the planes \(\overrightarrow{\mathbf{r}} \cdot(2 \hat{\mathbf{i}}-\lambda \hat{\mathbf{j}}+\hat{\mathbf{k}})=3\) and \(\overrightarrow{\mathbf{r}} \cdot(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mu \hat{\mathbf{k}})=5\) are parallel, then what is \(\lambda\) equal to?

1 \(1 / 2\)
2 1
3 \(-1 / 2\)
4 -1
SCRA-2012
Three Dimensional Geometry

121203 The Cartesian equation of the line passing through the point \((-1,3,-2)\) and perpendicular to the lines \(\quad \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}\) is

1 \(\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}\)
2 \(\frac{x-1}{-2}=\frac{y+3}{-7}=\frac{z-2}{-4}\)
3 \(\frac{x+1}{2}=\frac{y+3}{7}=\frac{z+2}{4}\)
4 \(\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry

121204 The equation of the plane, which bisects the line joining the points \((1,2,3)\) and \((3,4,5)\) at right angles is

1 \(x+y+z=0\)
2 \(x+y-z=0\)
3 \(x+y+z=9\)
4 \(\mathrm{x}+\mathrm{y}-\mathrm{z}+9=0\)
WB JEE-2019
For More Updates join with Us
Three Dimensional Geometry

121200 The equation of the line joining the points \((-3,4,11)\) and \((1,-2,7)\) is

1 \(\frac{x+3}{2}=\frac{y-4}{3}=\frac{z-11}{4}\)
2 \(\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}\)
3 \(\frac{x+3}{-2}=\frac{y+4}{3}=\frac{z+11}{4}\)
4 \(\frac{x+3}{2}=\frac{y+4}{-3}=\frac{z+11}{2}\)
Karnataka CET-2021
Three Dimensional Geometry

121201 If the planes \(\overrightarrow{\mathbf{r}} \cdot(2 \hat{\mathbf{i}}-\lambda \hat{\mathbf{j}}+\hat{\mathbf{k}})=3\) and \(\overrightarrow{\mathbf{r}} \cdot(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\mu \hat{\mathbf{k}})=5\) are parallel, then what is \(\lambda\) equal to?

1 \(1 / 2\)
2 1
3 \(-1 / 2\)
4 -1
SCRA-2012
Three Dimensional Geometry

121203 The Cartesian equation of the line passing through the point \((-1,3,-2)\) and perpendicular to the lines \(\quad \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}\) is

1 \(\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}\)
2 \(\frac{x-1}{-2}=\frac{y+3}{-7}=\frac{z-2}{-4}\)
3 \(\frac{x+1}{2}=\frac{y+3}{7}=\frac{z+2}{4}\)
4 \(\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry

121204 The equation of the plane, which bisects the line joining the points \((1,2,3)\) and \((3,4,5)\) at right angles is

1 \(x+y+z=0\)
2 \(x+y-z=0\)
3 \(x+y+z=9\)
4 \(\mathrm{x}+\mathrm{y}-\mathrm{z}+9=0\)
WB JEE-2019
NEET DIGITAL LIBRARY