Tags Maths

Maths

Get insights from 6.5k questions on Maths, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths

Follow Ask Question

6.5k

Questions

Discussions

Active Users

Followers

New answer posted

6 months ago

Maths 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

6 months ago

Maths Trigonometric Functions

22. Find the value of:

(i) sin 75° (ii) tan 15°

0 Follower 9 Views

Payal Gupta

Contributor-Level 10

22. (i) sin 75°= sin (45°+30°)

Using sin (x + y)= sin x cos y + cos x sin y we can write

sin 75°

= sin 45°cos 30°+ 45° sin 30°

0 0

New answer posted

6 months ago

Maths 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

6 months ago

Maths 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

6 months ago

Maths Trigonometric Functions

21. $2 s i n^{2} \frac{3 π}{4} + 2 c o s^{2} \frac{π}{4} + 2 s e c^{2} \frac{π}{3} = 10$

0 Follower 1 View

Payal Gupta

Contributor-Level 10

0 0

New answer posted

6 months ago

Maths 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

6 months ago

Maths 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

6 months ago

Maths Trigonometric Functions

20. $c o t^{2} \frac{π}{6} + c o s e c \frac{5 π}{6} + 3 t a n^{2} \frac{π}{6} = 6$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

0 0

New answer posted

6 months ago

Maths 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

6 months ago

Maths 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

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
679k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Maths

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience