Equation of a Line, Sphere, and a Plane in Different Forms
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Three Dimensional Geometry

121245 The equation of the line passing through the point \((2,3,-4)\) and perpendicular to \(\mathrm{XOZ}\) plane is

1 \(x=-2, y=-3+\lambda, z=4\)
2 \(\frac{x-2}{1}=\frac{z+4}{1}, y=3\)
3 \(x=-2, y=3+\lambda, z=4\)
4 \(x=2, y=3+\lambda, z=-4\)
MHT CET-2020
Three Dimensional Geometry

121247 If the lines \(\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}\) and \(\frac{x+1}{4}=\frac{1-3 y}{15}=z+1\) are perpendicular to each other, then \(\lambda=\)

1 3
2 5
3 4
4 2
MHT CET-2020
Three Dimensional Geometry

121248 The shortest distance between the Skew lines \(\overrightarrow{\mathbf{r}}=(3 \hat{i}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\lambda(-\hat{i}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \(\overrightarrow{\mathbf{r}}=(\hat{\mathbf{i}}-7 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) is

1 \(\frac{26}{5 \sqrt{5}}\)
2 \(\sqrt{45}\)
3 \(\sqrt{35}\)
4 \(\frac{36}{5 \sqrt{5}}\)
TS EAMCET-05.08.2021
Three Dimensional Geometry

121249 The foot of the perpendicular drawn from the origin to the plane \(x+y+3 z-4=0\) is

1 \(\left(\frac{2}{11}, \frac{2}{11}, \frac{9}{11}\right)\)
2 \(\left(\frac{1}{7}, \frac{1}{7}, \frac{6}{7}\right)\)
3 \(\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)\)
4 \(\left(\frac{1}{5}, \frac{1}{5}, \frac{3}{5}\right)\)
MHT CET-2020
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Three Dimensional Geometry

121245 The equation of the line passing through the point \((2,3,-4)\) and perpendicular to \(\mathrm{XOZ}\) plane is

1 \(x=-2, y=-3+\lambda, z=4\)
2 \(\frac{x-2}{1}=\frac{z+4}{1}, y=3\)
3 \(x=-2, y=3+\lambda, z=4\)
4 \(x=2, y=3+\lambda, z=-4\)
MHT CET-2020
Three Dimensional Geometry

121247 If the lines \(\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}\) and \(\frac{x+1}{4}=\frac{1-3 y}{15}=z+1\) are perpendicular to each other, then \(\lambda=\)

1 3
2 5
3 4
4 2
MHT CET-2020
Three Dimensional Geometry

121248 The shortest distance between the Skew lines \(\overrightarrow{\mathbf{r}}=(3 \hat{i}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\lambda(-\hat{i}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \(\overrightarrow{\mathbf{r}}=(\hat{\mathbf{i}}-7 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) is

1 \(\frac{26}{5 \sqrt{5}}\)
2 \(\sqrt{45}\)
3 \(\sqrt{35}\)
4 \(\frac{36}{5 \sqrt{5}}\)
TS EAMCET-05.08.2021
Three Dimensional Geometry

121249 The foot of the perpendicular drawn from the origin to the plane \(x+y+3 z-4=0\) is

1 \(\left(\frac{2}{11}, \frac{2}{11}, \frac{9}{11}\right)\)
2 \(\left(\frac{1}{7}, \frac{1}{7}, \frac{6}{7}\right)\)
3 \(\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)\)
4 \(\left(\frac{1}{5}, \frac{1}{5}, \frac{3}{5}\right)\)
MHT CET-2020
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Three Dimensional Geometry

121245 The equation of the line passing through the point \((2,3,-4)\) and perpendicular to \(\mathrm{XOZ}\) plane is

1 \(x=-2, y=-3+\lambda, z=4\)
2 \(\frac{x-2}{1}=\frac{z+4}{1}, y=3\)
3 \(x=-2, y=3+\lambda, z=4\)
4 \(x=2, y=3+\lambda, z=-4\)
MHT CET-2020
Three Dimensional Geometry

121247 If the lines \(\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}\) and \(\frac{x+1}{4}=\frac{1-3 y}{15}=z+1\) are perpendicular to each other, then \(\lambda=\)

1 3
2 5
3 4
4 2
MHT CET-2020
Three Dimensional Geometry

121248 The shortest distance between the Skew lines \(\overrightarrow{\mathbf{r}}=(3 \hat{i}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\lambda(-\hat{i}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \(\overrightarrow{\mathbf{r}}=(\hat{\mathbf{i}}-7 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) is

1 \(\frac{26}{5 \sqrt{5}}\)
2 \(\sqrt{45}\)
3 \(\sqrt{35}\)
4 \(\frac{36}{5 \sqrt{5}}\)
TS EAMCET-05.08.2021
Three Dimensional Geometry

121249 The foot of the perpendicular drawn from the origin to the plane \(x+y+3 z-4=0\) is

1 \(\left(\frac{2}{11}, \frac{2}{11}, \frac{9}{11}\right)\)
2 \(\left(\frac{1}{7}, \frac{1}{7}, \frac{6}{7}\right)\)
3 \(\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)\)
4 \(\left(\frac{1}{5}, \frac{1}{5}, \frac{3}{5}\right)\)
MHT CET-2020
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Three Dimensional Geometry

121245 The equation of the line passing through the point \((2,3,-4)\) and perpendicular to \(\mathrm{XOZ}\) plane is

1 \(x=-2, y=-3+\lambda, z=4\)
2 \(\frac{x-2}{1}=\frac{z+4}{1}, y=3\)
3 \(x=-2, y=3+\lambda, z=4\)
4 \(x=2, y=3+\lambda, z=-4\)
MHT CET-2020
Three Dimensional Geometry

121247 If the lines \(\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}\) and \(\frac{x+1}{4}=\frac{1-3 y}{15}=z+1\) are perpendicular to each other, then \(\lambda=\)

1 3
2 5
3 4
4 2
MHT CET-2020
Three Dimensional Geometry

121248 The shortest distance between the Skew lines \(\overrightarrow{\mathbf{r}}=(3 \hat{i}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\lambda(-\hat{i}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \(\overrightarrow{\mathbf{r}}=(\hat{\mathbf{i}}-7 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) is

1 \(\frac{26}{5 \sqrt{5}}\)
2 \(\sqrt{45}\)
3 \(\sqrt{35}\)
4 \(\frac{36}{5 \sqrt{5}}\)
TS EAMCET-05.08.2021
Three Dimensional Geometry

121249 The foot of the perpendicular drawn from the origin to the plane \(x+y+3 z-4=0\) is

1 \(\left(\frac{2}{11}, \frac{2}{11}, \frac{9}{11}\right)\)
2 \(\left(\frac{1}{7}, \frac{1}{7}, \frac{6}{7}\right)\)
3 \(\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)\)
4 \(\left(\frac{1}{5}, \frac{1}{5}, \frac{3}{5}\right)\)
MHT CET-2020
For More Updates join with Us