Maths

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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35. Let f (x) = cos (sin x).

f' (x) $\frac{d}{d x}$ cos (sin x)

= - sin (sin x) $\frac{d}{d x}$ sin x

= - sin (sin x) cos x.

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22. Find the shortest distance between the lines whose vector equations are

$\begin{array}{l} \vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} = (3 - 2 t) \hat{k} a n d \\ \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} = (2 s + 1) \hat{k} \end{array}$

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$\begin{array}{l} \vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k} \\ \vec{r} = \hat{i} - t \hat{i} + t \hat{j} - 2 \hat{j} + 3 \hat{k} - 2 t \hat{k} \\ \vec{r} = \hat{i} - 2 \hat{j} + 3 \hat{k} + t (- \hat{i} + \hat{j} - 2 \hat{k}) - - - - - (1) \end{array}$

$\begin{array}{l} \Rightarrow \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k} \\ \vec{r} = s \hat{i} + \hat{i} + 2 s \hat{j} - \hat{j} - 2 s \hat{k} - \hat{k} \\ \vec{r} = \hat{i} - \hat{j} - \hat{k} + s (\hat{i} + 2 \hat{j} - 2 \hat{k}) - - - - - (2) \end{array}$

Here,

Hence, the shortage distance between the line is given by

$d = | \frac{({\vec{b}}_{1} * {\vec{b}}_{2}) . ({\vec{a}}_{2} - {\vec{a}}_{1})}{| {\vec{b}}_{1} * {\vec{b}}_{2} |} | = 8$

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21. Find the shortest distance between the lines whose vector equations are

$\begin{array}{l} \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + 3 \hat{j} + 2 \hat{k}) \\ a n d \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k}) \end{array}$

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$\begin{array}{l} \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) a n d - - - - - (1) \\ \vec{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k}) - - - - - (2) \end{array}$

Here, comparing (1) and (2) with $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ , we have

$\begin{array}{l} a_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}, b_{1} = \hat{i} - 3 \hat{j} + 2 \hat{k} \\ a_{2} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}, b_{2} = 2 \hat{i} + 3 \hat{j} + \hat{k} \end{array}$

Therefore,

\begin{array}{l} {\vec{a}}_{2} - {\vec{a}}_{1} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ {\vec{b}}_{1} * {\vec{b}}_{2} = \\ = \hat{i} (- 3 - 6) - \hat{j} (1 - 4) + \hat{k} (3 + 6) \\ = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k} \\ | {\vec{b}}_{1} * {\vec{b}}_{2} | = = \\ = \\ = 3 \end{array}

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34. Kindly Consider the following

$s i n (x^{2} + 5)$

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34. Let f (x) = sin (x2 + 5)

Differentiating w. r t. x we get,

f'¢ (x) = $\frac{d}{d x}$ sin (x² + 5)

= cos (x² + 5) $\frac{d}{d x}$ = cos (x² + 5)

= cos (x² + 5) [2x].

= 2x cos (x² + 5).

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20. Find the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} a n d \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$ .

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$\begin{array}{l} \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \\ \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \end{array}$

Shortest distance between two lines is given by,

given equation we have

$\begin{array}{l} x_{1} = - 1, y_{1} = - 1, z_{1} = - 1 \\ a_{1} = 7, b_{1} = - 6, c_{1} = 1 \\ x_{2} = 3, y_{2} = 5, z_{2} = 7 \\ a_{2} = 1, b_{2} = - 2, c_{2} = 1 \end{array}$

Then,

$\begin{array}{l} = 4 (- 6 + 2) - 6 (7 - 1) + 8 (- 14 + 6) \\ = - 16 - 36 - 64 \\ = - 116 \end{array}$

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19. Find the shortest distance between the $\vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} + \hat{j} + \hat{k}) a n d \vec{r} = 2 \hat{i} + \hat{j} + \hat{k} + μ (2 \hat{i} + \hat{j} + 2 \hat{k})$ lines

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$\begin{array}{l} \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) a n d - - - - - (1) \\ \vec{r} = 2 \hat{i} - \hat{j} - \hat{k} + μ (2 \hat{i} + \hat{j} + 2 \hat{k}) - - - - - (2) \end{array}$

Solution. Comparing (1) and (2) with $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ respectively.

We get,

$\begin{array}{l} a_{1} = \hat{i} + 2 \hat{j} + \hat{k}, b_{1} = \hat{i} - \hat{j} + \hat{k} \\ a_{2} = 2 \hat{i} - \hat{j} - \hat{k}, b_{2} = 2 \hat{i} + \hat{j} + 2 \hat{k} \end{array}$

Therefore,

$\begin{array}{l} {\vec{a}}_{2} - {\vec{a}}_{1} = \hat{i} - 3 \hat{j} - 2 \hat{k} \\ {\vec{b}}_{1} * {\vec{b}}_{2} = (\hat{i} - \hat{j} + \hat{k}) * (2 \hat{i} + \hat{j} + 2 \hat{k}) \end{array}$

$\begin{array}{l} = (2 - 1) \hat{i} - (2 - 2) \hat{j} + (1 + 2) \hat{k} \\ = - 3 \hat{i} + 3 \hat{k} \\ | {\vec{b}}_{1} * {\vec{b}}_{2} | = = = = 3 \\ ({\vec{b}}_{1} * {\vec{b}}_{2}) . ({\vec{a}}_{2} - {\vec{a}}_{1}) = (- 3 \hat{i} + 3 \hat{k}) (\hat{i} - 3 \hat{j} - 2 \hat{k}) \\ = - 3 - 6 \\ = - 9 \end{array}$

Hence, the shortest distance between the given line is given by

$\begin{array}{l} d = | \frac{({\vec{b}}_{1} * {\vec{b}}_{2}) . ({\vec{a}}_{2} - {\vec{a}}_{1})}{| {\vec{b}}_{1} * {\vec{b}}_{2} |} | \\ = | \frac{- 9}{3} | \\ = \frac{3}{} = \frac{3}{2} \end{array}$

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18. Show that the lines $\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} a n d \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.

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$\begin{array}{l} \frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \\ \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \end{array}$

Direction ratios of given lines are (7, -5,1) and (1,2,3).

i.e., $\begin{array}{l} a_{1} = 7, b_{1} = - 5, c_{1} = 1 \\ a_{2} = 1, b_{2} = 2, c_{2} = 3 \end{array}$

Now,

$\begin{array}{l} = a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} \\ = 7 * 1 + (- 5) * 2 + 1 * 3 \\ = 7 - 10 + 3 \\ = 10 - 10 \\ = 0 \end{array}$

$∴$ These two lines are perpendicular to each other.

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33. Find all the points of discontinuity of $f$ defined by $f x x x$

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