PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).

Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\]  Reduce the corresponding equation in cartesian from.

Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]

The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]  Find a vector equation for the line.

Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are  \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\]  Also, reduce the equation obtained in vector form.

Find the direction cosines of the line  \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\]  Also, reduce it to vector form. 

The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form. 

Find the vector equation of a line passing through the point with position vector  \[\hat{i} - 2 \hat{j} - 3 \hat{k}\]  and parallel to the line joining the points with position vectors  \[\hat{i} - \hat{j} + 4 \hat{k} \text{ and } 2 \hat{i} + \hat{j} + 2 \hat{k} .\] Also, find the cartesian equivalent of this equation.

Find the points on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\]  at a distance of 5 units from the point P (1, 3, 3).

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.

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

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.

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.

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. 

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).

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).

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

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. 

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). 

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