Commerce (English Medium) कक्षा १२ - CBSE Question Bank Solutions

Find the equation of the plane which contains two parallel lines\[\frac{x - 4}{1} = \frac{y - 3}{- 4} = \frac{z - 2}{5}\text{  and }\frac{x - 3}{1} = \frac{y + 2}{- 4} = \frac{z}{5} .\]

Show that the lines  \[\frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2}\] and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.

Show that the plane whose vector equation is \[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j} - \hat{k}  \right) = 3\] contains the line whose vector equation is \[\vec{r} = \hat{i} + \hat{j}  + \lambda\left( 2 \hat{i}  + \hat{j} + 4 \hat{k}  \right) .\]

Find the equation of the plane determined by the intersection of the lines \[\frac{x + 3}{3} = \frac{y}{- 2} = \frac{z - 7}{6} \text{ and  }\frac{x + 6}{1} = \frac{y + 5}{- 3} = \frac{z - 1}{2}\]

Find the coordinates of the point where the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\] intersect the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.

Find the vector equation of the plane passing through three points with position vectors  \[\hat{i}  + \hat{j}  - 2 \hat{k}  , 2 \hat{i}  - \hat{j}  + \hat{k}  \text{ and }  \hat{i}  + 2 \hat{j}  + \hat{k}  .\]  Also, find the coordinates of the point of intersection of this plane and the line  \[\vec{r} = 3 \hat{i}  - \hat{j}  - \hat{k}  + \lambda\left( 2 \hat{i}  - 2 \hat{j} + \hat{k} \right) .\]

Find the distance of the point with position vector

The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y − z = −5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is

The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] is 

A plane meets the coordinate axes at A, B and C such that the centroid of ∆ABC is the point (a, b, c). If the equation of the plane is \[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = k,\] then k = 

A vector parallel to the line of intersection of the planes\[\vec{r} \cdot \left( 3 \hat{i} - \hat{j} + \hat{k}  \right) = 1 \text{ and }  \vec{r} \cdot \left( \hat{i} + 4 \hat{j}  - 2 \hat{k}  \right) = 2\] is 

The distance of the point (−1, −5, −10) from the point of intersection of the line \[\vec{r} = 2 \hat{i}- \hat{j} + 2 \hat{k}  + \lambda\left( 3 \hat{i}  + 4 \hat{j}+ 12 \hat{k}  \right)\]   and the plane \[\vec{r} \cdot \left( \hat{i} - \hat{j} + \hat{k}  \right) = 5\] is 

The equation of the plane through the intersection of the planes ax + by + cz + d = 0 andlx + my + nz + p = 0 and parallel to the line y=0, z=0

The equation of the plane which cuts equal intercepts of unit length on the coordinate axes is

Find the equation of the plane passing through the intersection of the planes `vecr . (hati + hatj + hatk)` and `vecr.(2hati + 3hatj - hatk) + 4 = 0` and parallel to the x-axis. Hence, find the distance of the plane from the x-axis.

Find the angle between the vectors `vec"a" + vec"b" and  vec"a" -vec"b" if  vec"a" = 2hat"i"-hat"j"+3hat"k" and vec"b" = 3hat"i" + hat"j"-2hat"k", and"hence find a vector perpendicular to both"  vec"a" + vec"b" and vec"a" - vec"b"`.

Show that the lines `("x"-1)/(3) = ("y"-1)/(-1) = ("z"+1)/(0) = λ and  ("x"-4)/(2) = ("y")/(0) = ("z"+1)/(3)` intersect. Find their point of intersection. 

The negative of a matrix is obtained by multiplying it by ______.

Evaluate the following:

`int sqrt(1 + x^2)/x^4 "d"x`

Evaluate the following:

`int ("d"x)/sqrt(16 - 9x^2)`