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Science (English Medium) इयत्ता १२ - CBSE Question Bank Solutions for Mathematics

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Mathematics
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Show that the points whose position vectors are  \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k}  \text{ and }  7 \text{ i}  - \text{ k} \]  are collinear.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line  \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\] 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Show that the three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular. 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 5).

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by  \[\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} .\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text { and }\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\]  are perpendicular to each other. 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1). 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the equation of a line parallel to x-axis and passing through the origin.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line: 

\[\overrightarrow{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) \text{ and }\overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line: 

\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k}  \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line: 

\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]

 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line:

\[\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} \text  { and }  \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line:

\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and } \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line:

\[\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} \text{  and  } \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line:

\[\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 \text{ and } \frac{x + 1}{1} = \frac{2y - 3}{3} = \frac{z - 5}{2}\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line:

\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{  and  } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the angle between the following pair of line:

\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{  and  } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined
< prev  3341 to 3360 of 8966  next > 
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