Commerce (English Medium) Class 12 - CBSE Question Bank Solutions for Mathematics

Show that the lines \[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \text{           and                  } \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}\]   intersect. Find their point of intersection.

Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection. 

Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.

Determine whether the following pair of lines intersect or not: 

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

Determine whether the following pair of lines intersect or not: 

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

Determine whether the following pair of lines intersect or not: 

\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]

Determine whether the following pair of lines intersect or not:  

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

Show that the lines \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j}  + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.

Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]

Find the perpendicular distance of the point (1, 0, 0) from the line  \[\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}.\] Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3). 

A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D. 

Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the line.

Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line  \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} .\]  Also, write down the coordinates of the foot of the perpendicular from P. 

Find the length of the perpendicular drawn from the point (5, 4, −1) to the line \[\overrightarrow{r} = \hat{i}  + \lambda\left( 2 \hat{i} + 9 \hat{j} + 5 \hat{k} \right) .\]

Find the foot of the perpendicular drawn from the point  \[\hat{i} + 6 \hat{j} + 3 \hat{k} \]  to the line  \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k}  \right) .\]  Also, find the length of the perpendicular

Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line  \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k}  \right) .\]  Also, find the coordinates of the foot of the perpendicular from P.

Find the foot of the perpendicular from (0, 2, 7) on the line \[\frac{x + 2}{- 1} = \frac{y - 1}{3} = \frac{z - 3}{- 2} .\]

Find the foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]

Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD.