Arts (English Medium) कक्षा १२ - CBSE Question Bank Solutions for Mathematics

If the lines  \[\frac{x - 1}{- 3} = \frac{y - 2}{- 2k} = \frac{z - 3}{2} \text{ and }\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}\] are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines.

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line  \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\] .

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

Find the image of the point with position vector \[3 \hat{i} + \hat{j}  + 2 \hat{k} \]  in the plane  \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j}  + \hat{k}  \right) = 4 .\]  Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through \[3 \hat{i}  + \hat{j}  + 2 \hat{k} .\]

Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to  \[\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .\]

Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3x − y − z = 7. Also, find the length of the perpendicular.

Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane \[\vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 4 \hat{k}  \right) + 5 = 0 .\]

Find the coordinates of the foot of the perpendicular and the perpendicular distance of the  point P (3, 2, 1) from the plane 2x − y + z + 1 = 0. Also, find the image of the point in the plane.

Find the direction cosines of the unit vector perpendicular to the plane  \[\vec{r} \cdot \left( 6 \hat{i}  - 3 \hat{j} - 2 \hat{k} \right) + 1 = 0\] passing through the origin.

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.

Find the length and the foot of perpendicular from the point \[\left( 1, \frac{3}{2}, 2 \right)\]  to the plane \[2x - 2y + 4z + 5 = 0\] .

Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained

Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector \[2 \hat{i}  + 3 \hat{j}  + 4 \hat{k} \] to the plane  \[\vec{r} . \left( 2 \hat{i} + \hat{j}  + 3 \hat{k}  \right) - 26 = 0\] Also find image of P in the plane.

Find the distance of the point P (–1, –5, –10) from the point of intersection of the line joining the points A (2, –1, 2) and B (5, 3, 4) with the plane x – y + z = 5.