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 Continuity and Differentiability

40. Kindly consider the following

0 Follower 1 View

alok kumar singh

Contributor-Level 10

40. Kindly go through the solution

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

27. Find the vector and Cartesian equations of the planes

(a) That passes through the point $(1, 0, - 2)$ and the normal to the plane is $\hat{i} - \hat{j} - \hat{k}$ .

(b) That passes through the point $(1, 4, 6)$ and the normal vector to the plane is $\hat{i} - 2 \hat{j} + \hat{k}$ .

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

(a) The position vector of point $(1, 0, - 2)$ is $\vec{a} = \hat{i} - 2 \hat{k}$

The normal vector N perpendicular to the plane is $\vec{N} = \hat{i} + \hat{j} - \hat{k}$

The vector equation of the plane is given by, $(\vec{r} - \vec{a}) . \vec{N} = 0$

$\Rightarrow [\vec{r} - (\hat{i} - 2 \hat{k})] . (\hat{i} + \hat{j} - \hat{k}) = 0 . . . . . . . . . (1)$

$\vec{r}$ is the position vector of any point $(x, y, z)$ in the plane.

$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Therefore, equation (1) becomes

$\begin{array}{l} [(x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} - 2 \hat{k})] . (\hat{i} + \hat{j} - \hat{k}) = 0 \\ \Rightarrow [(x - 1) \hat{i} + y \hat{j} + (z + 2) \hat{k}] . (\hat{i} + \hat{j} - \hat{k}) = 0 \\ \Rightarrow (x - 1) + y - (z + 2) = 0 \\ \Rightarrow x + y - z - 3 = 0 \\ \Rightarrow x + y - z = 3 \end{array}$

This is the Cartesian equation of the required plane.

(b) The position vector of the point $(1, 4, 6)$ is $\vec{a} = \hat{i} + 4 \hat{j} + 6 \hat{k}$

The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N} = \hat{i} - 2 \hat{j} + \hat{k}$

The vector equation of the plane is given by, $(\vec{r} - \vec{a}) . \vec{N} = 0$

$\Rightarrow [\vec{r} - (\hat{i} + 4 \hat{j} + 6 \hat{k})] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 . . . . . . . . . (1)$

$\vec{r}$ is the position vector of any point $P (x, y, z)$ in the plane.

$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Therefore, equation (1) becomes

$\begin{array}{l} [(x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} + 4 \hat{j} + 6 \hat{k})] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 \\ \Rightarrow [(x - 1) \hat{i} + (y - 4) \hat{j} + (z - 6) \hat{k}] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 \\ \Rightarrow (x - 1) + 2 (y - 4) + (z - 6) = 0 \\ \Rightarrow x - 2 y + z + 1 = 0 \end{array}$

This is the Car

...more

(a) The position vector of point $(1, 0, - 2)$ is $\vec{a} = \hat{i} - 2 \hat{k}$

The normal vector N perpendicular to the plane is $\vec{N} = \hat{i} + \hat{j} - \hat{k}$

The vector equation of the plane is given by, $(\vec{r} - \vec{a}) . \vec{N} = 0$

$\Rightarrow [\vec{r} - (\hat{i} - 2 \hat{k})] . (\hat{i} + \hat{j} - \hat{k}) = 0 . . . . . . . . . (1)$

$\vec{r}$ is the position vector of any point $(x, y, z)$ in the plane.

$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.

(b) The position vector of the point $(1, 4, 6)$ is $\vec{a} = \hat{i} + 4 \hat{j} + 6 \hat{k}$

The normal vector $\vec{N}$ perpendicular to the plane is $\vec{N} = \hat{i} - 2 \hat{j} + \hat{k}$

The vector equation of the plane is given by, $(\vec{r} - \vec{a}) . \vec{N} = 0$

$\Rightarrow [\vec{r} - (\hat{i} + 4 \hat{j} + 6 \hat{k})] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 . . . . . . . . . (1)$

$\vec{r}$ is the position vector of any point $P (x, y, z)$ in the plane.

$∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.

less

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Continuity and Differentiability

39. Kindly consider the following

$c o s x^{3} s i n^{2} (x^{5})$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

39. Let f (x) = cos (x³) sin² (x⁵).

f' (x) = cos (x³) $\frac{d}{d x}$ sin² (x⁵) + sin² (x⁵) $\frac{d}{d x}$ cos (x³)

= cos (x³) 2sin (x⁵) $\frac{d}{d x}$ sin (x⁵) + sin² (x⁵) [sin (x³)] $\frac{d}{d x}$ x³.

= 2 cos (x³) sin (x⁵). cos (x⁵) $\frac{d}{d x}$ (x⁵) - sin² (x⁵) sin (x³). 3x²

= 2. cos (x³) sin (x⁵) cos (x⁵). 5 - 3x²sin² (x⁵) sin (x³)

= x² sin (x⁵). [2x² cos (x³) cos (x⁵) - 3 sin (x⁵) sin x³].

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

26. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin:

$\begin{array}{l} \end{array} (a) 2 x + 3 y + 4 z - 12 = 0$

$(b) 3 y + 4 z - 6 = 0$

$(c) x + y + z = 1$

$(d) 5 y + 8 = 0$

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be $(x_{1}, y_{1}, z_{1}) .$

$\begin{array}{l} 2 x + 3 y + 4 z - 12 = 0 \end{array}$

$\begin{array}{l} \Rightarrow 2 x + 3 y + 4 z = 12 \dots (1) \end{array}$

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

25. Find the Cartesian equation of the following planes:

$\begin{array}{l} (a) \vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2 \\ (b) \vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1 \\ (c) \vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \end{array}$

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value

...more

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$\begin{array}{l} (x \hat{i} + y \hat{j} - z \hat{k}) . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \\ \Rightarrow (s - 2 t) x + (3 - t) y + (2 s + t) z = 15 \end{array}$

This is the Cartesian equation of the given plane.

less

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Continuity and Differentiability

38. Kindly consider the following

$\frac{s i n (a x + b)}{c o s (c x + d)}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

38. Let f(x) = $\frac{s i n (a x + b)}{c o s (c x + d)} .$

f'(x) = $\frac{c o s (c x + d) \frac{d}{d x} s i n (a x + b) - s i n (a x + b) \frac{d}{d x} c o s (c x + d)}{c o s^{2} (c x + d) .}$

= $\frac{c o s (c x + d) c o s (a x + b) \frac{d}{d x} (a x + b) + s i n (a x + b) s i n (c x + d) \frac{d (x + d)}{d x}}{c o s^{2} (c x + d)}$

= $= \frac{c o s (c x + d) c o s (a x + b) \cdot a + s i n (a x + b) s i n (c x + d) \cdot c}{c o s^{2} (c x + d)}$

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Continuity and Differentiability

37. Kindly consider the following

0 Follower 1 View

alok kumar singh

Contributor-Level 10

37. Kindly go through the solution

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

24. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value

...more

(a) It is given that equation of the plane is

$\vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ I s given by, $\vec{r} = x \hat{i} + y \hat{j} - z \hat{k}$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2$

$\Rightarrow x + y - z = 2$

This is the Cartesian equation of the plane.

(b) $\vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$(x \hat{i} + y \hat{j} - z \hat{k}) . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1$

$\Rightarrow 2 x + 3 y - 4 z = 1$

This is the Cartesian equation of the plane.

(c) $\vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15$

For any arbitrary point $P (x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r} = (x \hat{i} + y \hat{j} - z \hat{k})$

Substituting the value of $\vec{r}$ in equation (1), we obtain

$\begin{array}{l} (x \hat{i} + y \hat{j} - z \hat{k}) . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \\ \Rightarrow (s - 2 t) x + (3 - t) y + (2 s + t) z = 15 \end{array}$

This is the Cartesian equation of the given plane.

less

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

23. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

$(a$ $)$ $z = 2$

$(b) x + y + z = 1$

$(c) 2 x + 3 y - z = 5$

$(d) 5 y + 8 = 0$

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

(a) The equation of the plane is $z = 2 o r 0 x + 0 y + z = 2 . . . . . . . . . . (1)$

The direction ratios of normal are $0, 0, a n d 1 .$

$∴ + 02 + 12 = 1$

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

$0 . x + 0 . y + 1 . z = 2$

This is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x + y + z = 1 . . . . . . . . . . (1)$

The direction ratios of normal are 1, 1, and 1.

$∴ + (1) ² + (1) ² =$

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

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{} . . . . . . . . . (2)$

This equation is of the form $l x + m y + n z = d$ , where l, m, n

...more

(a) The equation of the plane is $z = 2 o r 0 x + 0 y + z = 2 . . . . . . . . . . (1)$

The direction ratios of normal are $0, 0, a n d 1 .$

$∴ + 02 + 12 = 1$

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

$0 . x + 0 . y + 1 . z = 2$

This is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x + y + z = 1 . . . . . . . . . . (1)$

The direction ratios of normal are 1, 1, and 1.

$∴ + (1) ² + (1) ² =$

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

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{} . . . . . . . . . (2)$

This equation is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are $\frac{1}{}$ , $\frac{1}{}$ and $\frac{1}{}$ the distance of normal from the origin is $\frac{1}{}$ units.

(c) $2 x + 3 y - z = 5 . . . . . . . . . . (1)$

The direction ratios of normal are 2, 3, and −1.

$∴ + (3) ² + (- 1) ² =$

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

$\frac{2}{x} + \frac{3}{y} - \frac{1}{z} = \frac{5}{}$

This equation is of the form $l x + m y + n z = d$ where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane $\frac{2}{x} + \frac{3}{y} - \frac{1}{z}$ and the distance of normal from the origin is $\frac{5}{}$ units.

(d)

$\begin{array}{l} 5 y + 8 = 0 & \Rightarrow 0 x - 5 y + 0 z = 8 . . . . . . . . . . (1) \end{array}$

The direction ratios of normal are 0, −5, and 0.

$∴ + (- 5) ² + 0 = 5$

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

$- y = \frac{8}{5}$

This equation is of the form $l x + m y + n z = d$ where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin is $\frac{8}{5}$ units.

less

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Continuity and Differentiability

36. Kindly consider the following

$s i n (a x + b)$

0 Follower 1 View

alok kumar singh

Contributor-Level 10

36. Let f (x) = sin (ax + b)

f' (x) = $\frac{d}{d x}$ sin (ax + b)

= cos (ax + b) $\frac{d}{d x}$ (ax + b)

= a cos (ax + b).

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