Class 12th

41. Kindly consider the following

0 Follower 1 View

Contributor-Level 10

41. Kindly consider the following

0 0

New answer posted

8 months ago

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 7 Views

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

New answer posted

8 months ago

40. Kindly consider the following

0 Follower 1 View

Contributor-Level 10

40. Kindly go through the solution

0 0

New answer posted

8 months ago

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 3 Views

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

8 months ago

39. Kindly consider the following

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

0 Follower 2 Views

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

8 months ago

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

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

8 months ago

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

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

8 months ago

38. Kindly consider the following

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

0 Follower 3 Views

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

8 months ago

37. Kindly consider the following

0 Follower 2 Views

Contributor-Level 10

37. Kindly go through the solution

0 0

New answer posted

8 months ago

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