#### Online Mock Tests

#### Chapters

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solution of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

## Chapter 29: The Plane

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.1 [Pages 4 - 5]

Find the equation of the plane passing through the following points.

(2, 1, 0), (3, −2, −2) and (3, 1, 7)

Find the equation of the plane passing through the following points.

(−5, 0, −6), (−3, 10, −9) and (−2, 6, −6)

Find the equation of the plane passing through the following point

(1, 1, 1), (1, −1, 2) and (−2, −2, 2)

Find the equation of the plane passing through the following points.

(2, 3, 4), (−3, 5, 1) and (4, −1, 2)

Find the equation of the plane passing through the following point

(0, −1, 0), (3, 3, 0) and (1, 1, 1)

Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane.

Show that the following points are coplanar.

(0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0)

Show that the following points are coplanar.

(0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1)

Find the coordinates of the point *P *where the line through *A *(3, -4 , -5 ) and B (2, -3 , 1) crosses the plane passing through three points *L*(2,2,1), M(3,0,1) and N(4, -1,0 ) . Also, find the ratio in which *P *diveides the line segment *AB.*

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.2 [Page 7]

Write the equation of the plane whose intercepts on the coordinate axes are 2, −3 and 4.

Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes.

4*x* + 3*y* − 6*z* − 12 = 0

Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes.

2x + 3y − z = 6

Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes.

2*x* − *y* + *z* = 5

Find the equation of a plane which meets the axes at *A*, *B* and *C*, given that the centroid of the triangle *ABC* is the point (α, β, γ).

Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinate axes.

A plane meets the coordinate axes at *A*, *B* and *C,* respectively, such that the centroid of triangle *ABC* is (1, −2, 3). Find the equation of the plane.

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.3 [Pages 13 - 14]

Find the vector equation of a plane passing through a point with position vector \[2 \hat{i} - \hat{j} + \hat{k} \] and perpendicular to the vector \[4 \hat{i} + 2 \hat{j} - 3 \hat{k} .\]

Find the Cartesian form of the equation of a plane whose vector equation is

\[\vec{r} \cdot \left( 12 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + 5 = 0\]

Find the Cartesian form of the equation of a plane whose vector equation is

\[\vec{r} \cdot \left( - \hat{i} + \hat{j} + 2 \hat{k} \right) = 9\]

Find the vector equations of the coordinate planes.

Find the vector equation of each one of following planes.

2*x* − *y* + 2*z* = 8

Find the vector equation of each one of following planes.

*x* + *y* − *z* = 5

Find the vector equation of each one of following planes.

*x* + *y* = 3

Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).

\[\vec{n}\] is a vector of magnitude \[\sqrt{3}\] and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to \[\vec{n}\] .

The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.

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.

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*.

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

A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point

Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

Show that the normals to the following pairs of planes are perpendicular to each other.

*x* − *y* + *z* − 2 = 0 and 3*x* + 2*y* − *z* + 4 = 0

Show that the normals to the following pairs of planes are perpendicular to each other.

Show that the normal vector to the plane 2*x* + 2*y* + 2*z* = 3 is equally inclined to the coordinate axes.

Find a vector of magnitude 26 units normal to the plane 12*x* − 3*y* + 4*z* = 1.

If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.

Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.

Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.

If *O* be the origin and the coordinates of *P* be (1, 2,−3), then find the equation of the plane passing through *P* and perpendicular to *OP*.

Find the vector equation of the plane with intercepts 3, –4 and 2 on *x*, *y* and *z*-axis respectively.

Find the vector equation of the plane with intercepts 3, –4 and 2 on *x*, *y* and *z*-axis respectively.

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.4 [Page 19]

Find the vector equation of a plane which is at a distance of 3 units from the origin and has \[\hat{k}\] as the unit vector normal to it.

Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector \[\hat{i} - \text{2 } \hat{j} - \text{2 } \hat{k} .\]

Reduce the equation 2*x* − 3*y* − 6*z* = 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.

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

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

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 equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane

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 distance of the plane 2*x* − 3*y* + 4*z* − 6 = 0 from the origin.

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.5 [Pages 22 - 23]

Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5).

Find the vector equation of the plane passing through the points *P* (2, 5, −3), *Q* (−2, −3, 5) and *R* (5, 3, −3).

Find the vector equation of the plane passing through points *A* (*a*, 0, 0), *B* (0, *b*, 0) and *C*(0, 0, *c*). Reduce it to normal form. If plane *ABC* is at a distance *p* from the origin, prove that \[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .\]

Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).

Find the vector equation of the plane passing through the points \[3 \hat{i} + 4 \hat{j} + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k} \text{ and } 7 \hat{i} + 6 \hat{k} .\]

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.6 [Page 29]

Find the angle between the given planes. \[\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) = 1 \text{ and } \vec{r} \cdot \left( - \hat{i} + \hat{j} \right) = 4\]

Find the angle between the given planes. \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 6 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + 6 \hat{j} - 2 \hat{k} \right) = 9\]

Find the angle between the planes.

2*x* − *y* + *z* = 4 and *x* + *y* + 2*z* = 3

Find the angle between the planes.

*x* + *y* − 2*z* = 3 and 2*x* − 2*y* + *z* = 5

Find the angle between the planes.

*x* − *y* + *z* = 5 and *x* + 2*y* + *z* = 9

Find the angle between the planes.

2*x* − 3*y* + 4*z* = 1 and − *x* + *y* = 4

Find the angle between the planes.

2*x* + *y* − 2*z* = 5 and 3*x* − 6*y* − 2*z* = 7

Show that the following planes are at right angles.

\[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( - \hat{i} - \hat{j} + \hat{k} \right) = 3\]

Show that the following planes are at right angles.

*x* − 2*y* + 4*z* = 10 and 18*x* + 17*y* + 4*z* = 49

Determine the value of λ for which the following planes are perpendicular to each other.

Determine the value of λ for which the following planes are perpendicular to each ot

2*x* − 4*y* + 3*z* = 5 and *x* + 2*y* + λ*z* = 5

Determine the value of λ for which the following planes are perpendicular to each other.

3*x* − 6*y* − 2*z* = 7 and 2*x* + *y* − λ*z* = 5

Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3*x* + 2*y* − 3*z* = 1 and 5*x* − 4*y* + *z* = 5.

Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes *x* + 2*y* + 2*z* = 5 and 3*x* + 3*y* + 2*z* = 8.

Find the equation of the plane passing through the origin and perpendicular to each of the planes *x* + 2*y* − *z* = 1 and 3*x* − 4*y* + *z* = 5.

Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6*x* − 2*y* + 2*z* = 9.

Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2*x* + 6*y* + 6*z* = 1.

Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane *x* + 2*y* + 2*z* = 5.

Find the equation of the plane with intercept 3 on the *y*-axis and parallel to the *ZOX *plane.

Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2*x* + 3*y* − 2*z* = 5 and *x* + 2*y* − 3*z* = 8.

Find the equation of the plane passing through (*a*, *b*, *c*) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes *x* + 2*y* + 3*z* = 5 and 3*x* + 3*y* + *z* = 0.

Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane *x* − 2*y* + 4*z* = 10

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.7 [Page 33]

Find the vector equations of the following planes in scalar product form \[\left( \vec{r} \cdot \vec{n} = d \right):\] \[\vec{r} = \left( 2 \hat{i} - \hat{k} \right) + \lambda \hat{i} + \mu\left( \hat{i} - 2 \hat{j} - \hat{k}

\right)\]

Find the vector equations of the following planes in scalar product form \[\left( \vec{r} \cdot \vec{n} = d \right):\] \[\vec{r} = \left( 1 + s - t \right) \hat{t} + \left( 2 - s \right) \hat{j} + \left( 3 - 2s + 2t \right) \hat{k} \]

Find the vector equations of the following planes in scalar product form \[\left( \vec{r} \cdot \vec{n} = d \right):\]\[\vec{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - \hat{k} \right) + \mu\left( - \hat{i} + \hat{j} - 2 \hat{k} \right)\]

Find the vector equations of the following planes in scalar product form \[\left( \vec{r} \cdot \vec{n} = d \right):\]\[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \right)\]

Find the Cartesian forms of the equations of the following planes. \[\vec{r} = \left( \hat{i} - \hat{j} \right) + s\left( - \hat{i} + \hat{j} + 2 \hat{k} \right) + t\left( \hat{i} + 2 \hat{j} + \hat{k} \right)\]

Find the Cartesian forms of the equations of the following planes.

Find the vector equation of the following planes in non-parametric form. \[\vec{r} = \left( \lambda - 2\mu \right) \hat{i} + \left( 3 - \mu \right) \hat{j} + \left( 2\lambda + \mu \right) \hat{k} \]

Find the vector equation of the following planes in non-parametric form. \[\vec{r} = \left( 2 \hat{i} + 2 \hat{j} - \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \mu\left( 5 \hat{i} - 2 \hat{j} + 7 \hat{k} \right)\]

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.8 [Pages 39 - 40]

Find the equation of the plane which is parallel to 2*x* − 3*y* + *z* = 0 and which passes through (1, −1, 2).

Find the equation of the plane through (3, 4, −1) which is parallel to the plane \[\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j} + 5 \hat{k} \right) + 2 = 0 .\]

Find the equation of the plane passing through the line of intersection of the planes 2*x* − 7*y* + 4*z* − 3 = 0, 3*x* − 5*y* + 4*z* + 11 = 0 and the point (−2, 1, 3).

Find the equation of the plane through the point \[2 \hat{i} + \hat{j} - \hat{k} \] and passing through the line of intersection of the planes \[\vec{r} \cdot \left( \hat{i} + 3 \hat{j} - \hat{k} \right) = 0 \text{ and } \vec{r} \cdot \left( \hat{j} + 2 \hat{k} \right) = 0 .\]

Find the equation of the plane passing through the line of intersection of the planes 2*x* − *y* = 0 and 3*z* − *y* = 0 and perpendicular to the plane 4*x* + 5*y* − 3*z* = 8

Find the equation of the plane which contains the line of intersection of the planes *x* + 2*y* + 3*z* − 4 = 0 and 2*x* + *y* − *z* + 5 = 0 and which is perpendicular to the plane 5*x* + 3*y* − 6*z*+ 8 = 0.

Find the equation of the plane through the line of intersection of the planes *x* + 2*y* + 3*z* + 4 = 0 and *x* − *y* + *z* + 3 = 0 and passing through the origin.

Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes *x* − 3*y* + 2*z* − 5 = 0 and 2*x* − *y* + 3*z* − 1 = 0 and passing through (1, −2, 3).

Find the equation of the plane that is perpendicular to the plane 5*x* + 3*y* + 6*z* + 8 = 0 and which contains the line of intersection of the planes *x* + 2*y* + 3*z* − 4 = 0, 2*x* + *y* − *z* + 5 = 0.

Find the equation of the plane through the line of intersection of the planes \[\vec{r} \cdot \left( \hat{i} + 3 \hat{j} \right) + 6 = 0 \text{ and } \vec{r} \cdot \left( 3 \hat{i} - \hat{j} - 4 \hat{k} \right) = 0,\] which is at a unit distance from the origin.

Find the equation of the plane passing through the intersection of the planes 2*x* + 3*y* − *z*+ 1 = 0 and *x* + *y* − 2*z* + 3 = 0 and perpendicular to the plane 3*x* − *y* − 2*z* − 4 = 0.

Find the equation of the plane that contains the line of intersection of the planes \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) - 4 = 0 \text{ and } \vec{r} \cdot \left( 2 \hat{i} + \hat{j} - \hat{k} \right) + 5 = 0\] and which is perpendicular to the plane \[\vec{r} \cdot \left( 5 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) + 8 = 0 .\]

Find the equation of the plane passing through (*a*, *b*, *c*) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]

Find the equation of the plane passing through the intersection of the planes \[\vec{r} \cdot \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) = 7, \vec{r} \cdot \left( 2 \hat{i} + 5 \hat{j} + 3 \hat{k} \right) = 9\] and the point (2, 1, 3).

Find the equation of the plane through the intersection of the planes 3*x* − *y* + 2*z* = 4 and *x* + *y* + *z* = 2 and the point (2, 2, 1).

Find the vector equation of the plane through the line of intersection of the planes *x* + *y*+ *z* = 1 and 2*x* + 3*y* + 4*z* = 5 which is perpendicular to the plane *x* − *y* + *z* = 0.

Find the vector equation of the plane passing through the intersection of the planes

\[\vec{r} \cdot \left( \hat{ i } + \hat{ j }+ \hat{ k }\right) = \text{ 6 and }\vec{r} \cdot \left( \text{ 2 } \hat{ i} +\text{ 3 } \hat{ j } + \text{ 4 } \hat{ k } \right) = - 5\] and the point (1, 1, 1).

Find the equation of the plane which contains the line of intersection of the planes *x *\[+\] 2y \[+\] 3 \[z - \] 4 \[=\] 0 and 2 \[x + y - z\] \[+\] 5 \[=\] 0 and whose *x*-intercept is twice its *z*-intercept. Hence, write the equation of the plane passing through the point (2, 3, \[-\] 1) and parallel to the plane obtained above.

Find the equation of the plane through the line of intersection of the planes \[x + y + z =\]1 and 2*x *\[+\] 3 \[+\] y \[+\] 4\[z =\] 5 and twice of its \[y\] *-*intercept is equal to three times its \[z\]-intercept

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.9 [Page 49]

Find the distance of the point \[2 \hat{i} - \hat{j} - 4 \hat{k}\] from the plane \[\vec{r} \cdot \left( 3 \hat{i} - 4 \hat{j} + 12 \hat{k} \right) - 9 = 0 .\]

Show that the points \[\hat{i} - \hat{j} + 3 \hat{k} \text{ and } 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \] are equidistant from the plane \[\vec{r} \cdot \left( 5 \hat{i} + 2 \hat{j} - 7 \hat{k} \right) + 9 = 0 .\]

Find the distance of the point (2, 3, −5) from the plane *x* + 2*y* − 2*z* − 9 = 0.

Find the equations of the planes parallel to the plane *x* + 2*y* − 2*z* + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1).

Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3*x* + 4*y* − 12*z* + 13 = 0.

Find the equations of the planes parallel to the plane *x* − 2*y* + 2*z* − 3 = 0 and which are at a unit distance from the point (1, 1, 1).

Find the distance of the point (2, 3, 5) from the *xy *- plane.

Find the distance of the point (3, 3, 3) from the plane \[\vec{r} \cdot \left( 5 \hat{i} + 2 \hat{j} - 7k \right) + 9 = 0\]

If the product of the distances of the point (1, 1, 1) from the origin and the plane *x* − *y* + *z*+ λ = 0 be 5, find the value of λ.

Find an equation for the set of all points that are equidistant from the planes 3*x* − 4*y* + 12*z* = 6 and 4*x* + 3*z* = 7.

Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C (5, 3, −3).

A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.

Find the distance of the point (1, -2, 4) from plane passing throuhg the point (1, 2, 2) and perpendicular of the planes x - y + 2z = 3 and 2x - 2y + z + 12 = 0

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.1 [Page 51]

Find the distance between the parallel planes 2*x* − *y* + 3*z* − 4 = 0 and 6*x* − 3*y* + 9*z* + 13 = 0.

Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2*x* − 3*y* + 5*z* + 7 = 0. Also, find the distance between the two planes.

Find the equation of the plane mid-parallel to the planes 2*x* − 2*y* + *z* + 3 = 0 and 2*x* − 2*y* + *z* + 9 = 0.

Find the distance between the planes \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + 7 = 0 \text{ and } \vec{r} \cdot \left( 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \right) + 7 = 0 .\]

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.11 [Pages 61 - 62]

Find the angle between the line \[\vec{r} = \left( 2 \hat{i}+ 3 \hat {j} + 9 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right)\] and the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 5 .\]

Find the angle between the line \[\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z + 1}{1}\] and the plane 2*x* + *y* − *z* = 4.

Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3*x* − *y* + *z* = 1.

The line \[\vec{r} = \hat{i} + \lambda\left( 2 \hat{i} - m \hat{j} - 3 \hat{k} \right)\] is parallel to the plane \[\vec{r} \cdot \left( m \hat{i} + 3 \hat{j} + \hat{k} \right) = 4 .\] Find *m*.

Show that the line whose vector equation is \[\vec{r} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k}+ \lambda\left( \hat{i} + 3 \hat{j} + 4 \hat{k} \right)\] is parallel to the plane whose vector \[\vec{r} \cdot \left( \hat{i} + \hat{j} - \hat{k} \right) = 7 .\] Also, find the distance between them.

Find the vector equation of the line through the origin which is perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 3 .\]

Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to *x*-axis.

Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line \[\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7} .\]

Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes \[\vec{r} \cdot \left( \hat{i} - \hat{j} + 2 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j} + 2 \hat{k} \right) = 6\]

Prove that the line of section of the planes 5*x* + 2*y* − 4*z* + 2 = 0 and 2*x* + 8*y* + 2*z* − 1 = 0 is parallel to the plane 4*x* − 2*y* − 5*z* − 2 = 0.

Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2*x* − *y* + 3*z* − 5 = 0.

Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

Find the angle between the line \[\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{2}\] and the plane

3*x* + 4*y* + *z* + 5 = 0.

Find the equation of the plane passing through the intersection of the planes *x* − 2*y* + *z* = 1 and 2*x* + *y* + *z* = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane

State when the line \[\vec{r} = \vec{a} + \lambda \vec{b}\] is parallel to the plane \[\vec{r} \cdot \vec{n} = d .\]Show that the line \[\vec{r} = \hat{i} + \hat{j} + \lambda\left( 3 \hat{i} - \hat{j} + 2 \hat{k} \right)\] is parallel to the plane \[\vec{r} \cdot \left( 2 \hat{j} + \hat{k} \right) = 3 .\] Also, find the distance between the line and the plane.

Show that the plane whose vector equation is \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - \hat{k} \right) = 1\] and the line whose vector equation is \[\vec{r} = \left( - \hat{i} + \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 4 \hat{k} \right)\] are parallel. Also, find the distance between them.

Find the equation of the plane through the intersection of the planes 3*x* − 4*y* + 5*z* = 10 and 2*x* + 2*y* − 3*z* = 4 and parallel to the line *x* = 2*y* = 3*z*.

Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines \[\vec{r} = \left( \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)\] and \[\vec{r} = \left( \hat{i} - 3 \hat{j} + 5 \hat{k} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\] Also, find the distance of the point (9, −8, −10) from the plane thus obtained.

Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line

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

Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - 5 \hat{k} \right) + 9 = 0 .\]

Find the angle between the line

*x*+ 2

*y*− 11

*z*= 3.

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes \[\vec{r} \cdot \left( \hat{i} - \hat{j} + 2 \hat{k} \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j} + \hat{k} \right) = 6 .\]

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 3*x* − *y* − 2*z* = 7.

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

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.12 [Page 65]

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the *yz *- plane .

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the *zx *- plane .

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

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

Find the distance of the point (2, 12, 5) from the point of intersection of the line \[\vec{r} = 2 \hat{i} - 4 \hat{j}+ 2 \hat{k} + \lambda\left( 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \right)\] and \[\vec{r} . \left( \hat{i} - 2 \hat{j} + \hat{k} \right) = 0\]

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 \[x - y + z = 5\] .

Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2*x* + *y* + *z* = 7.

Find the distance of the point (1, -5, 9) from the plane

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.13 [Pages 73 - 74]

Show that the lines \[\vec{r} = \left( 2 \hat{j} - 3 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) \text{ and } \vec{r} = \left( 2 \hat{i} + 6 \hat{j} + 3 \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right)\] are coplanar. Also, find the equation of the plane containing them.

Show that the lines \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} \text{ and }\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\] are coplanar. Also, find the equation of the plane containing them.

Find the equation of the plane containing the line \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}\] and the point (0, 7, −7) and show that the line \[\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\] also lies in the same plane.

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 3*x* − 2*y* + *z* + 5 = 0 = 2*x* + 3*y* + 4*z* − 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 vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2*x* − 5*y* − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j} - 2 \hat{k} + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) .\]

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 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 \[\vec{r} = 3 \hat{i} - \hat{j} - \hat{k} + \lambda\left( 2 \hat{i} - 2 \hat{j} + \hat{k} \right) .\]

Show that the lines \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5}\] and \[\frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.

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

Show that the lines \[\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}\] and \[\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}\] are coplanar. Hence, find the equation of the plane containing these lines.

If the line \[\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z + 4}{3}\] lies in the plane \[lx + my - z =\] then find the value of \[l^2 + m^2\] .

Find the values of \[\lambda\] for which the lines

If the lines \[x =\] 5 , \[\frac{y}{3 - \alpha} = \frac{z}{- 2}\] and \[x = \alpha\] \[\frac{y}{- 1} = \frac{z}{2 - \alpha}\] are coplanar, find the values of \[\alpha\].

If the straight lines \[\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}\] and \[\frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z}{k}\] are coplanar, find the equations of the planes containing them.

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.14 [Page 77]

Find the shortest distance between the lines

Find the shortest distance between the lines

Find the shortest distance between the lines

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.15 [Pages 81 - 82]

Find the image of the point (0, 0, 0) in the plane 3*x* + 4*y* − 6*z* + 1 = 0.

Find the reflection of the point (1, 2, −1) in the plane 3*x* − 5*y* + 4*z* = 5.

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

Hence, or otherwise, deduce the length of the perpendicular.

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 coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2*x* − 2*y* + 4*z* + 5 = 0. Also, find the length of the perpendicular.

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 3*x* − *y* − *z* = 7. Also, find the length of the perpendicular.

Find the image of the point (1, 3, 4) in the plane 2*x* − *y* + *z* + 3 = 0.

Find the distance of the point with position vector

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 2*x* − *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 2*x* − 3*y* + 4*z* − 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 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.

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane Very Short Answers [Pages 83 - 84]

Write the equation of the plane parallel to *XOY*- plane and passing through the point (2, −3, 5).

Write the equation of the plane parallel to the *YOZ*- plane and passing through (−4, 1, 0).

Write the equation of the plane passing through points (*a*, 0, 0), (0, *b*, 0) and (0, 0, *c*).

Write the general equation of a plane parallel to *X*-axis.

Write the value of *k* for which the planes *x* − 2*y* + *kz* = 4 and 2*x* + 5*y* − *z* = 9 are perpendicular.

Write the intercepts made by the plane 2*x* − 3*y* + 4*z* = 12 on the coordinate axes.

Write the ratio in which the plane 4*x* + 5*y* − 3*z* = 8 divides the line segment joining the points (−2, 1, 5) and (3, 3, 2).

Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.

Write the plane \[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) = 14\] in normal form.

Write the distance of the plane \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 12\] from the origin.

Write the equation of the plane \[\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}\] in scalar product form.

Write a vector normal to the plane \[\vec{r} = l \vec{b} + m \vec{c} .\]

Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3*x* + 2*y* −*z* = 7.

Write the equation of the plane containing the lines \[\vec{r} = \vec{a} + \lambda \vec{b} \text{ and } \vec{r} = \vec{a} + \mu \vec{c} .\]

Write the position vector of the point where the line \[\vec{r} = \vec{a} + \lambda \vec{b}\] meets the plane \[\vec{r} . \vec{n} = 0 .\]

Write the value of *k* for which the line \[\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}\] is perpendicular to the normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) = 4 .\]

Write the angle between the line \[\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{- 2}\] and the plane *x* + *y* + 4 = 0.

Write the intercept cut off by the plane 2*x* + *y* − *z* = 5 on *x*-axis.

Find the length of the perpendicular drawn from the origin to the plane 2*x* − 3*y* + 6*z* + 21 = 0.

Write the vector equation of the line passing through the point (1, −2, −3) and normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right) = 5 .\]

Find the vector equation of the plane, passing through the point (*a*, *b*, *c*) and parallel to the plane \[\vec{r} . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2\]

Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \[2 \hat{i} - 3 \hat{j} + 6 \hat{k} \] .

Write the equation of a plane which is at a distance of \[5\sqrt{3}\] units from origin and the normal to which is equally inclined to coordinate axes.

### RD Sharma solutions for Class 12 Maths Chapter 29 The Plane MCQ [Pages 84 - 86]

The plane 2*x* − (1 + λ) *y* + 3λ*z* = 0 passes through the intersection of the planes

2

*x*−*y*= 0 and*y*− 3*z*= 02

*x*+ 3*z*= 0 and*y*= 02

*x*−*y*+ 3*z*= 0 and*y*− 3*z*= 0None of these

The acute angle between the planes 2*x* − *y* + *z* = 6 and *x* + *y* + 2*z* = 3 is

45°

60°

30°

75°

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

7

*x*− 2*y*+ 3*z*+ 81 = 023

*x*+ 14*y*− 9*z*+ 48 = 051

*x*− 15*y*− 50*z*+ 173 = 0None of these

The distance between the planes 2*x* + 2*y* − *z* + 2 = 0 and 4*x* + 4*y* − 2*z* + 5 = 0 is

\[\frac{1}{2}\]

\[\frac{1}{4}\]

\[\frac{1}{6}\]

None of these

The image of the point (1, 3, 4) in the plane 2*x* − *y* + *z* + 3 = 0 is

(3, 5, 2)

(−3, 5, 2)

(3, 5, −2)

(3, −5, 2)

The equation of the plane containing the two lines

8

*x*+*y*− 5*z*− 7 = 08

*x*+*y*+ 5*z*− 7 = 08

*x*−*y*− 5*z*− 7 = 0None of these

The equation of the plane \[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - 2 \hat{j} + 3 \hat{k} \right)\] in scalar product form is

\[\vec{r} \cdot \left( 5 \hat{i} - 2 \hat{j} - 3 \hat{k} \right) = 7\]

\[\vec{r} \cdot \left( 5 \hat{i} + 2 \hat{j} - 3 \hat{k} \right) = 7\]

\[\vec{r} \cdot \left( 5 \hat{i} - 2 \hat{j} + 3 \hat{k} \right) = 7\]

None of these

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

\[\frac{5}{3\sqrt{3}}\]

\[\frac{10}{3\sqrt{3}}\]

\[\frac{25}{3\sqrt{3}}\]

None of these

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

*x*− 5*y*+ 3*z*= 7*x*− 5*y*+ 3*z*= −7*x*+ 5*y*+ 3*z*= 7*x*+ 5*y*+ 3*z*= −7

The vector equation of the plane containing the line \[\vec{r} = \left( - 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) + \lambda\left( 3 \hat{i} - 2 \hat{j} - \hat{k} \right)\] and the point \[\hat{i} + 2 \hat{j} + 3 \hat{k} \] is

\[\vec{r} \cdot \left( \hat{i} + 3 \hat{k} \right) = 10\]

\[\vec{r} \cdot \left( \hat{i} - 3 \hat{k} \right) = 10\]

\[\vec{r} \cdot \left( 3\hat{i} - \hat{k} \right) = 10\]

None of these

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 =

1

2

3

None of these

*x*+

*y*+

*z*= 17 is

1

2

3

None of these

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

\[- 2 \hat{i} + 7 \hat{j}+ 13 \hat{k} \]

\[2 \hat{i} + 7 \hat{j} - 13 \hat{k}\]

\[-2 \hat{i} + 7 \hat{j} + 13 \hat{k}\]

\[2 \hat{i} + 7 \hat{j} + 13 \hat{k}\]

If a plane passes through the point (1, 1, 1) and is perpendicular to the line \[\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}\] then its perpendicular distance from the origin is

3/4

4/3

7/5

1

The equation of the plane parallel to the lines *x* − 1 = 2*y* − 5 = 2*z* and 3*x* = 4*y* − 11 = 3*z* − 4 and passing through the point (2, 3, 3) is

*x*− 4*y*+ 2*z*+ 4 = 0*x*+ 4*y*+ 2*z*+ 4 = 0*x*− 4*y*+ 2*z*− 4 = 0None of these

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

9

13

17

None of these

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

(

*bl*−*am*)*y*+ (*cl*−*an*)*z*+*dl*−*ap*= 0(

*am*−*bl*)*x*+ (*mc*−*bn*)*z*+*md*−*bp*= 0(

*na*−*cl*)*x*+ (*bn*−*cm*)*y*+*nd*−*cp*= 0None of these

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

*x*+*y*+*z*= 1*x*+*y*+*z*= 0*x*+*y*−*z*= 1*x*+*y*+*z*= 2

## Chapter 29: The Plane

## RD Sharma solutions for Class 12 Maths chapter 29 - The Plane

RD Sharma solutions for Class 12 Maths chapter 29 (The Plane) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 12 Maths solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 12 Maths chapter 29 The Plane are Three - Dimensional Geometry Examples and Solutions, Introduction of Three Dimensional Geometry, Equation of a Plane Passing Through Three Non Collinear Points, Relation Between Direction Ratio and Direction Cosines, Intercept Form of the Equation of a Plane, Coplanarity of Two Lines, Distance of a Point from a Plane, Angle Between Line and a Plane, Angle Between Two Planes, Angle Between Two Lines, Vector and Cartesian Equation of a Plane, Equation of a Plane in Normal Form, Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point, Plane Passing Through the Intersection of Two Given Planes, Shortest Distance Between Two Lines, Equation of a Line in Space, Direction Cosines and Direction Ratios of a Line.

Using RD Sharma Class 12 solutions The Plane exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 12 prefer RD Sharma Textbook Solutions to score more in exam.

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