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

Let `veca = hati + hatj + hatk = hati` and `vecc = c_1veci + c_2hatj + c_3hatk` then

1) Let `c_1 = 1` and `c_2 = 2`, find `c_3` which makes `veca, vecb "and" vecc`coplanar

2) if `c_2 = -1` and `c_3 = 1`, show that no value of `c_1`can make `veca, vecb and vecc` coplanar

if `A = ((2,3,1),(1,2,2),(-3,1,-1))`, Find `A^(-1)` and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8

Give a condition that three vectors \[\vec{a}\], \[\vec{b}\] and \[\vec{c}\]  form the three sides of a triangle. What are the other possibilities?

Prove that a necessary and sufficient condition for three vectors \[\vec{a}\], \[\vec{b}\], \[\vec{c}\]  to be coplanar is that there exist scalars l, m, n not all zero simultaneously such that \[l \vec{a} + m \vec{b} + n \vec{c} = \vec{0} .\]

If |A| = 3 and \[A^{- 1} = \begin{bmatrix}3 & - 1 \\ - \frac{5}{3} & \frac{2}{3}\end{bmatrix}\] , then write the adj A .

Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.

Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. 

Reduce the equation \[\vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right) + 6 = 0\] to normal form and, hence, find the length of the perpendicular from the origin to the plane.

The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane. 

Find the equation of a plane which is at a distance of \[3\sqrt{3}\]  units from the origin and the normal to which is equally inclined to the coordinate axes.

Find the vector equation of the plane which is at a distance of \[\frac{6}{\sqrt{29}}\] from the origin and its normal vector from the origin is  \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} .\] Also, find its Cartesian form. 

Find the equation of the plane which contains the line of intersection of the planes \[x + 2y + 3z - 4 = 0 \text { and } 2x + y - z + 5 = 0\] and whose x-intercept is twice its z-intercept.

Find the value of λ such that the line \[\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}\]  is perpendicular to the plane 3x − y − 2z = 7.

Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line \[\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{- 1}\]