Maths

38. $\frac{c o s 4 x + c o s 3 x + c o s 2 x}{s i n 4 x + s i n 3 x + s i n 2 x} = c o t 3 x$

0 Follower 2 Views

Contributor-Level 10

38. L.H.S $= \frac{c o s 4 x + c o s 3 x + c o s 2 x}{s i n 4 x + s i n 3 x + s i n 2 x .}$

$= \frac{(c o s 4 x + c o s 2 x) + c o s 3 x}{(s i n 4 x + s i n 2 x) + s i n 3 x} .$

$= \frac{2 c o s (\frac{4 x + 2 x}{2}) c o s (\frac{4 x - 2 x}{2}) + c o s 3 x}{2 s i n (\frac{4 x + 2 x}{2}) c o s (\frac{4 x - 2 x}{2}) s i n 3 x} .$

$= \frac{2 c o s \frac{6 x}{2} c o s \frac{2 x}{2} + c o s 3 x}{2 s i n \frac{6 x}{2} c o s \frac{2 x}{2} + s i n 3 x}$

$= \frac{2 c o s 3 x c o s x + c o s 3 x}{2 s i n 3 x c o s x + 3 x}$

$= \frac{c o s 3 x}{s i n 3 x} * \frac{(2 c o s x + 1)}{(2 c o s x + 1)}$

$= c o t 3 x =R.H.S$

0 0

New answer posted

6 months ago

59. The planes: $2 x - y + 4 z = 5 a n d 5 x - 2.5 y + 10 z = 6$ are

(A) Perpendicular (B) Parallel (C) intersect y-axis

(D) Passes through (0,0,5/4)

0 Follower 2 Views

Contributor-Level 10

The equations of the planes are

$\begin{array}{l} 2 x - y + 4 z = 5 \dots (1) & 5 x - 2.5 y + 10 z = 6 \dots (2) \end{array}$

It can be seen that,

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{2}{5} \\ \frac{b_{1}}{b_{2}} = \frac{- 1}{- 2.5} = \frac{2}{5} \\ \frac{c_{1}}{c_{2}} = \frac{4}{10} = \frac{2}{5} \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Therefore, the given planes are parallel.

Hence, the correct answer is B.

0 0

New answer posted

6 months ago

Choose the correct answer in Exercise Q. 22 and 23.

58. Distance between the two planes: $2 x + 3 y + 4 z = 4 a n d 4 x + 6 y + 8 z = 12$ is

(A) 2 units (B) 4 units (C) 8 units (D) 2/√29 units

0 Follower 5 Views

Contributor-Level 10

The equations of the planes are

$\begin{array}{l} 2 x + 3 y + 4 z = 4 & 4 x + 6 y + 8 z = 12 & \Rightarrow 2 x + 3 y + 4 z = 6 \end{array}$

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes, $a x + b y + c z = d_{1} a n d a x + b y + c z = d_{2},$ is given by,

$\begin{array}{l} D = | \frac{d_{2} - d_{1}}{} | \\ \Rightarrow D = | \frac{6 - 4}{} | \\ D = \frac{2}{} \end{array}$

Thus, the distance between the lines is 2/√29 units.

Hence, the correct answer is D.

0 0

New answer posted

6 months ago

37. $\frac{s i n x - s i n 3 x}{s i n^{2} x - c o s^{2} x} = 2 s i n x$

0 Follower 2 Views

Contributor-Level 10

37. $= \frac{s i n x - s i n 3 x}{s i n^{2} x - c o s^{2} x}$

$= \frac{2 c o s (\frac{x + 3 x}{2}) s i n \frac{x - 3 x}{2}}{- (c o s^{2} x - s i n^{2} x)}$

$= \frac{2 c o s \frac{4 x}{2} s i n (- \frac{2 x}{2})}{- c o s 2 x} [? c o s 2 x = c o s^{2} x - s i n^{2} x]$

$= - \frac{2 c o s 2 x s i n x}{- c o s 2 x} [? s i n (- x) = - s i n x]$

= 2 sin x

= R.H.S.

0 0

New answer posted

6 months ago

57. Prove that if a plane has the intercepts a,b,c and is at a distance of P units from the origin, then $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$

0 Follower 2 Views

Contributor-Level 10

The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

The distance (p) of the plane from the origin is given by,

$\begin{array}{l} p = | \frac{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}{} | \\ \Rightarrow p = \frac{1}{} \\ \Rightarrow p^{2} = \frac{1}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} \\ \Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} \end{array}$

0 0

New answer posted

6 months ago

36. $\frac{s i n x + s i n 3 x}{c o s x + c o s 3 x} = t a n 2 x$

0 Follower 2 Views

Contributor-Level 10

36. $= \frac{s i n x + s i n 3 x}{c o s x + c o s 3 x}$

$= \frac{2 (\frac{x + 3 x}{2}) c o s (\frac{x - 3 x}{2})}{2 c o s (\frac{x + 3 x}{2}) c o s (\frac{x - 3 x}{2})}$

$= \frac{s i n \frac{4 x}{2}}{c o s \frac{4 x}{2}}$

$= \frac{s i n 2 x}{c o s 2 x}$

$= t a n 2 x =$ R.H.S

0 0

New answer posted

6 months ago

56. Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines:

$\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$

0 Follower 39 Views

Contributor-Level 10

Let the required line be parallel to the vector $\vec{b}$ given by, $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, − 4) is $\vec{a} = \hat{i} + 2 \hat{j} - 4 \hat{k}$

The equation of the line passing through (1, 2, −4) and parallel to vector $\vec{b}$ is

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . (1) \end{array}$

The equations of the lines are

$\begin{array}{l} \frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7} . . . . . . . . (2) \\ \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5} . . . . . . . . (3) \end{array}$

Line (1) and line (2) are perpendicular to each other.

$∴ 3 b_{1} - 16 b_{2} + 7 b_{3} = 0 . . . . . . . . (4)$

Also, line (1) and line (3) are perpendicular to each other.

$∴ 3 b_{1} + 8 b_{2} - 5 b_{3} = 0 . . . . . . . . (5)$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 16) (- 5) - 8 * 7} = \frac{b_{2}}{7 * 3 - 3 (- 5)} = \frac{b_{3}}{3 * 8 - 3 (- 16)} \\ \Rightarrow \frac{b_{1}}{24} = \frac{b_{2}}{36} = \frac{b_{3}}{72} \\ \Rightarrow \frac{b_{1}}{2} = \frac{b_{2}}{3} = \frac{b_{3}}{6} \end{array}$

Direction ratios of $\vec{b}$ are 2, 3, and 6.

$∴ \vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$

Substituting $\vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$

This is the equation of the required line.

0 0

New answer posted

6 months ago

35. $\frac{s i n x - s i n y}{c o s x + c o s y} = t a n \frac{x - y}{2}$

0 Follower 2 Views

Contributor-Level 10

35. L.H.S: = $L.H. S: = \frac{s i n x - s i n y}{c o s x + c o s y}$

$= \frac{2 c o s \frac{x + y}{2} s i n \frac{x - y}{2}}{2 c o s \frac{x + y}{2} c o s \frac{x - y}{2}}$

$= \frac{s i n (\frac{x - y}{2})}{c o s (\frac{x - y}{2})}$

$= t a n (\frac{x - y}{2}) = R.H.S.$ = R.H.S.

0 0

New answer posted

6 months ago

55. Find the vector equation of the line passing through (1, 2, 3) and parallel to the plane $\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5$ and $\vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6$

0 Follower 5 Views

Contributor-Level 10

Let the required line be parallel to vector $\vec{b}$ given by,

$\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, 3) is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The equation of line passing through (1, 2, 3) and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . . (1) \end{array}$

The equations of the given planes are

$\begin{array}{l} \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5 . . . . . . . . (2) \\ \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6 . . . . . . . . (3) \end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

$\begin{array}{l} \Rightarrow (\hat{i} - \hat{j} + 2 \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0 \\ \Rightarrow λ (b_{1} - b_{2} + 2 b_{3}) = 0 \\ \Rightarrow b_{1} - b_{2} + 2 b_{3} = 0 . . . . . . . . . . (4) \end{array}$

Similarly, $(3 \hat{i} + \hat{j} + \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 1) * 1 - 1 * 2} = \frac{b_{2}}{2 * 3 - 1 * 1} = \frac{b_{3}}{1 * 1 - 3 * (- 1)} \\ \Rightarrow \frac{b_{1}}{- 3} = \frac{b_{2}}{5} = \frac{b_{3}}{4} \end{array}$

Therefore, the direction ratios of $\vec{b}$ are −3, 5, and 4.

$∴ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = - 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$

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

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} + 5 \hat{j} + 4 \hat{k})$

This is the equat

...more

Let the required line be parallel to vector $\vec{b}$ given by,

$\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, 3) is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The equation of line passing through (1, 2, 3) and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . . (1) \end{array}$

The equations of the given planes are

$\begin{array}{l} \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5 . . . . . . . . (2) \\ \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6 . . . . . . . . (3) \end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

Similarly, $(3 \hat{i} + \hat{j} + \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 1) * 1 - 1 * 2} = \frac{b_{2}}{2 * 3 - 1 * 1} = \frac{b_{3}}{1 * 1 - 3 * (- 1)} \\ \Rightarrow \frac{b_{1}}{- 3} = \frac{b_{2}}{5} = \frac{b_{3}}{4} \end{array}$

Therefore, the direction ratios of $\vec{b}$ are −3, 5, and 4.

$∴ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = - 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$

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

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} + 5 \hat{j} + 4 \hat{k})$

This is the equation of the required line.

less

0 0

New answer posted

6 months ago

53. Find the equation of the plane which contains the line of intersection of the planes $\vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0, \vec{r} . (2 \hat{i} + h t \hat{j} - \hat{k}) + 5 = 0$ and which is perpendicular to the plane $\vec{r} . (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) + 8 = 0$

0 Follower 7 Views