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Arts (English Medium) Class 12 - CBSE Question Bank Solutions

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Find the coordinates of the point where the line through (3, ­−4, −5) and (2, − 3, 1) crosses the plane 2x + z = 7).

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

The planes: 2− y + 4z = 5 and 5x − 2.5y + 10z = 6 are

(A) Perpendicular

(B) Parallel

(C) intersect y-axis

(C) passes through `(0,0,5/4)`

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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Find the value of λ, if four points with position vectors `3hati + 6hatj+9hatk`, `hati + 2hatj + 3hatk`,`2hati + 3hatj + hatk` and `4hati + 6hatj + lambdahatk` are coplanar.

[10] Vectors
Chapter: [10] Vectors
Concept: undefined >> undefined

Find the coordinates of the point where the line through the points (3, - 4, - 5) and (2, - 3, 1), crosses the plane determined by the points (1, 2, 3), (4, 2,- 3) and (0, 4, 3)

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

If A is a 3 × 3 invertible matrix, then what will be the value of k if det(A–1) = (det A)k

[3] Matrices
Chapter: [3] Matrices
Concept: undefined >> undefined

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

[10] Vectors
Chapter: [10] Vectors
Concept: undefined >> undefined

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

[3] Matrices
Chapter: [3] Matrices
Concept: undefined >> undefined

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?

[10] Vectors
Chapter: [10] Vectors
Concept: undefined >> undefined

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 \[l \vec{a} + m \vec{b} + n \vec{c} = \vec{0} .\]

[10] Vectors
Chapter: [10] Vectors
Concept: undefined >> undefined

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

[3] Matrices
Chapter: [3] Matrices
Concept: undefined >> undefined

Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2.

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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. 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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.

 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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. 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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.

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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. 

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined
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