Class 12th

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 $\hat{i} - 2 \hat{j} + \hat{k}$ .

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(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

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(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.

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39. Kindly consider the following

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

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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].

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

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

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

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

$(d) 5 y + 8 = 0$

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

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

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

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11 months ago

25. Find the Cartesian equation of the following planes:

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

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(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

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(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.

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38. Kindly consider the following

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

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

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

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

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

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37. Kindly consider the following

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37. Kindly go through the solution

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

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(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

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(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.

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

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

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

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

$(d) 5 y + 8 = 0$

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(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

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(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.

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36. Kindly consider the following

$s i n (a x + b)$

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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).

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35. Kindly consider the following

$c o s (s i n x)$

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