Tags Ncert Solutions Maths class 12th

Ncert Solutions Maths class 12th

Get insights from 2.5k questions on Ncert Solutions Maths class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Maths class 12th

Follow Ask Question

2.5k

Questions

Discussions

Active Users

Followers

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

36. In the following cases, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them.

$\begin{array}{l} \end{array} (a) 7 x + 5 y + 6 z + 30 = 0 a n d 3 x - y - 10 z + 4 = 0$

$(b) 2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

$(c) 2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

$(d) 2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

$(e) 4 x + 8 y + z - 8 = 0 a n d y + z - 4 = 0$

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

The direction ratios of normal to the plane, $L_{1} : a_{1} x + b_{1} y + c_{1} z = 0 .$ , are $a_{1}, b_{1}, c_{1}$ and $L_{2} : a_{2} x + b_{2} y + c_{2} z = 0 .$

$L_{1}$
$\begin{array}{l} L_{2}, i f \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ L_{1} ⊥ L_{2}, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \end{array}$

The angle between $L_{1} & L_{2}$ is given by,

(b) The equations of the planes are $2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

$\begin{array}{l} a_{1} = 2, b_{1} = 1, c_{1} = 3 & a_{2} =, b_{2} = - 2, c_{2} = \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 1 + 1 * (- 2) + 3 * 0 = 0 \end{array}$

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

Here, $a_{1} = 2, b_{1} = - 2, c_{1} = 4 & a_{2} =, b_{2} = - 3, c_{2} = 6$

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 3 + 1 * (- 2) (- 3) + 4 * 6 = 6 + 6 + 24 = 36 \neq 0$

Thus, the given planes are not perpendicular to each other.

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{- 2}{- 3} = \frac{2}{3} & \frac{c_{1}}{c_{2}} = \frac{4}{6} = \frac{2}{3} \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Thus, the given planes are parallel to each other

(d) The equations of the planes are and $2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

$\begin{array}{l} a_{1} = 2, b_{1} = - 1, c_{1} = 3 & a_{2} =, b_{2} = - 1, c_{2} = 3 \\ \frac{a_{1}}{a_{2}} = \frac{2}{2} = 1, \frac{b_{1}}{b_{2}} = \frac{- 1}{- 1} = 1 & \frac{c_{1}}{c_{2}} = \frac{3}{3} = 1 \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Thus, the given lines are parallel to each other

(e) The equations of the given planes are $4 x + 8 y + z - 8 = 0 a n d y + z - 4 = 0$ $\begin{array}{l} a_{1} = 4, b_{1} = 8, c_{1} = 1 & a_{2} =, b_{2} = 1, c_{2} = 1 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 4 * 0 + 8 * 1 + 1 = 9 \neq 0 \end{array}$

Therefore, the given lines are not perpendicular to each

...more

The direction ratios of normal to the plane, $L_{1} : a_{1} x + b_{1} y + c_{1} z = 0 .$ , are $a_{1}, b_{1}, c_{1}$ and $L_{2} : a_{2} x + b_{2} y + c_{2} z = 0 .$

$L_{1}$
$\begin{array}{l} L_{2}, i f \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ L_{1} ⊥ L_{2}, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \end{array}$

The angle between $L_{1} & L_{2}$ is given by,

(b) The equations of the planes are $2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

$\begin{array}{l} a_{1} = 2, b_{1} = 1, c_{1} = 3 & a_{2} =, b_{2} = - 2, c_{2} = \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 1 + 1 * (- 2) + 3 * 0 = 0 \end{array}$

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

Here, $a_{1} = 2, b_{1} = - 2, c_{1} = 4 & a_{2} =, b_{2} = - 3, c_{2} = 6$

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 3 + 1 * (- 2) (- 3) + 4 * 6 = 6 + 6 + 24 = 36 \neq 0$

Thus, the given planes are not perpendicular to each other.

Thus, the given planes are parallel to each other

(d) The equations of the planes are and $2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

Thus, the given lines are parallel to each other

Therefore, the given lines are not perpendicular to each other.

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{4}{0}, \frac{b_{1}}{b_{2}} = \frac{8}{1} = 8 & \frac{c_{1}}{c_{2}} = \frac{1}{1} = 1 \\ ∴ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \end{array}$

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

less

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

35. Find the angle between the planes whose vector equations are $\vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 & \vec{r} . (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

The equations of the given planes are $\vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 & \vec{r} . (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$

It is known that if n1 and n2 are normal to the planes, $\vec{r} . \vec{n_{1}} = d_{1} & \vec{r} . \vec{n_{2}} = d_{2}$ then the angle between them, Q, is given by,

$c o s Q = | \frac{\vec{n_{1}} . \vec{n_{2}}}{| \vec{n_{1}} | | \vec{n_{2}} |} | . . . . . . . . . . (1)$

$\begin{array}{l} H e r e, \vec{n_{1}} = 2 \hat{i} + 2 \hat{j} - 3 \hat{k} & \vec{n_{2}} = 3 \hat{i} - 3 \hat{j} + 5 \hat{k} \\ ∴ \vec{n_{1}} . \vec{n_{2}} = (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 2.3 + 2 . (- 3) + (- 3) . 5 = - 15 \\ | \vec{n_{1}} | = = \\ | \vec{n_{2}} | = = \end{array}$

Substituting the value of $\vec{n_{1}} . \vec{n_{2}}$ $| \vec{n_{1}} | & | \vec{n_{2}} |$ in equation (1), we obtain

$\begin{array}{l} c o s Q = | \frac{- 15}{.} | \\ \Rightarrow c o s Q = \frac{15}{} \\ \Rightarrow c o s Q^{- 1} = (\frac{15}{}) \end{array}$

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

34. Find the equation of the plane through the line of intersection of the planes $x + y + z = 1 a n d 2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

The equation of the plane through the intersection of the planes, $x + y + z = 1 a n d 2 x + 3 y + 4 z = 5$ , is

$(x + y + z - 1) + λ (2 x + 3 y + 4 z - 5)$

$\Rightarrow (2 λ + 1) x + (3 λ + 1) y + (4 λ + 1) z - (5 λ + 1) = 0 . . . . . . . . . . (1)$

The direction ratios, $a_{1}, b_{1}, c_{1},$ of this plane are $(2 λ + 1), (3 λ + 1), a n d (4 λ + 1) .$

The plane in equation (1) is perpendicular to $x - y + z = 0$

Its direction ratios, $a_{2}, b_{2}, c_{2},$ are $1, - 1, a n d 1$ .

Since the planes are perpendicular,

$\begin{array}{l} a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \\ \Rightarrow (2 λ + 1) - (3 λ + 1) + (4 λ + 1) = 0 \\ \Rightarrow 3 λ + 1 = 0 \\ \Rightarrow λ = - \frac{1}{3} \end{array}$

Substituting $λ = - \frac{1}{3}$ in equation (1), we obtain

$\begin{array}{l} \frac{1}{3} x - \frac{1}{3} z + \frac{2}{3} = 0 \\ \Rightarrow x - z + 2 = 0 \end{array}$

This is the required equation of the plane.

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

33. Find the vector equation of the plane passing through the intersection of the planes $\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 7, \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$ and through the point $(2, 1, 3) .$

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

The equations of the planes are $\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 7, \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9$

$\begin{array}{l} \Rightarrow \vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) - 7 = 0 . . . . . . . . . . (1) \\ \Rightarrow \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9 = 0 . . . . . . . . . . (2) \end{array}$

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

$\begin{array}{l} [\vec{r} . (2 \hat{i} + 2 \hat{j} - 2 \hat{k}) - 7] + λ [\vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9] = 0, w h e r e, λ \in R \\ \vec{r} . [(2 \hat{i} + 2 \hat{j} - 2 \hat{k}) + λ (2 \hat{i} + 5 \hat{j} + 3 \hat{k})] = 9 λ + 7 \\ \vec{r} . [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} + (3 λ - 3) \hat{k}] = 9 λ + 7 . . . . . . . . . . (3) \end{array}$

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

$\vec{r} = 2 \hat{i} + 1 \hat{j} + 3 \hat{k}$

Substituting in equation (3), we obtain

$\begin{array}{l} (2 \hat{i} + \hat{j} + 3 \hat{k}) . [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} + (3 λ - 3) \hat{k}] = 9 λ + 7 \\ \Rightarrow 2 (2 + 2 λ) + 1 (2 + 5 λ) + 3 (3 λ - 3) = 9 λ + 7 \\ \Rightarrow 4 + 4 λ + 2 + 5 λ + 9 λ - 9 = 9 λ + 7 \\ \Rightarrow 18 λ - 3 = 9 λ + 7 \\ \Rightarrow 9 λ = 10 \\ \Rightarrow λ = \frac{10}{9} \end{array}$

Substituting $λ = \frac{10}{9}$ in equation (3), we obtain

$\begin{array}{l} \vec{r} . (\frac{38}{9} \hat{i} + \frac{68}{9} \hat{j} + \frac{3}{9} \hat{k}) = 17 \\ \Rightarrow \vec{r} . (38 \hat{i} + 68 \hat{j} + 3 \hat{k}) = 153 \end{array}$

This is the vector equation of the required plane.

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

31. Find the equation of the plane through the intersection of the planes $3 x - y + 2 z - 4 = 0 a n d x +$ $y + z - 2 = 0$ and the point $(2, 2, 1) .$

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

The equation of any plane through the intersection of the planes,

$3 x - y + 2 z - 4 = 0 a n d x + y + z - 2 = 0,$ is

$(3 x - y + 2 z - 4) + α (x + y + z - 2) = 0, w h e r e, α \in R . . . . . . . . . . (1)$

The plane passes through the point $(2, 2, 1) .$ Therefore, this point will satisfy equation (1).

$\begin{array}{l} ∴ (3 * 2 - 2 + 2 * 1 - 4) + α (2 + 2 + 1 - 2) = 0 \\ \Rightarrow 2 + 3 α = 0 \\ \Rightarrow α = - \frac{2}{3} \end{array}$

Substituting $α = - \frac{2}{3}$ in equation (1), we obtain

$\begin{array}{l} (3 x - y + 2 z - 4) - \frac{2}{3} (x + y + z - 2) = 0 \\ \Rightarrow 3 (3 x - y + 2 z - 4) - 2 (x + y + z - 2) = 0 \\ \Rightarrow (9 x - 3 y + 6 z - 12) - 2 (x + y + z - 2) = 0 \\ \Rightarrow 7 x - 5 y + 4 z - 8 = 0 \end{array}$

0 0

New question posted

4 months ago

Ncert Solutions Maths class 12th Continuity and Differentiability

42. Prove that the function $f$ given by

$f (x) = | x - 1 |, x \in$ is not differentiable at $x$ = 1.

0 Follower 3 Views

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

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

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

The equation of the plane ZOX is

y = 0

Any plane parallel to it is of the form, y = a

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

29. Find the intercepts cut off by the plane $2 x + y - z = 5$

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

$2 x + y - z = 5 . . . . . . . . . (1)$

Dividing both sides of equation (1) by 5, we obtain

$\begin{array}{l} \frac{2}{5} x + \frac{y}{5} - \frac{z}{5} = 1 \\ \Rightarrow \frac{x}{\frac{5}{2}} + \frac{y}{5} + \frac{z}{- 5} = 1 . . . . . . . . . . (2) \end{array}$

It is known that the equation of a plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ , where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively.

Therefore, for the given equation,

$a = \frac{5}{2}, b = 5 a n d c = - 5$

Thus, the intercepts cut off by the plane are $\frac{5}{2}$ $, 5 a n d - 5 .$

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Continuity and Differentiability

41. Kindly consider the following

0 Follower 1 View

alok kumar singh

Contributor-Level 10

41. Kindly consider the following

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

28. Find the equations of the planes that passes through three points:

(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)

(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

We know that through three collinear points $A, B, C$ i.e., through a straight line, we can pass an infinite number of planes.

(a) The given points are $A (1, 1, - 1), B (6, 4, - 5), a n d C (- 4, - 2, 3) .$

$\begin{array}{l} | \begin{matrix} 1 & 1 & - 1 \\ 6 & 4 & - 5 \\ - 4 & - 2 & 3 \end{matrix} | = (12 - 10) - (18 - 20) - (- 12 + 16) \end{array}$

$\begin{array}{l} = 2 + 2 - 4 \\ = 0 \end{array}$

Since $A, B, C$ are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are $A (1, 1, 0), B (1, 2, 1), a n d C (- 2, 2, - 1) .$

$| \begin{matrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ - 2 & 2 & - 1 \end{matrix} | = (- 2 - 2) - (2 + 2) = - 8 \neq 0$

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points, $(x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) & (x_{3}, y_{3}, z_{3})$ , is

$\begin{array}{l} | \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} \\ x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ x_{3} - x_{1} & y_{3} - y_{1} & z_{3} - z_{1} \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} x - 1 & y - 1 & z \\ 0 & 1 & 1 \\ - 3 & 1 & - 1 \end{matrix} | = 0 \\ \Rightarrow (- 2) (x - 1) - 3 (y - 1) + 3 z = 0 \\ \Rightarrow - 2 x - 3 y + 3 z + 2 + 3 = 0 \\ \Rightarrow - 2 x - 3 y + 3 z = - 5 \\ \Rightarrow 2 x + 3 y - 3 z = 5 \end{array}$

This is the Cartesian equation of the required plane.

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Ncert Solutions Maths class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience