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Maths Trigonometric Functions

29. sin²6x – sin²4x = sin 2x sin 10x

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Payal Gupta

Contributor-Level 10

29. L.H.S $L.H.S = s i n^{2} 6 x - s i n^{2} 4 x .$

$= (s i n 6 x + s i n 4 x) (s i n 6 x - s i n 4 x) . [a^{2} - b^{2} = (a - b) (a + b)]$

So, L.H.S $L.H.S. = (2 s i n (\frac{6 x + 4 x}{2}) c o s (\frac{6 x - 4 x}{2})) (2 c o s (\frac{6 x + 4 x}{2}) s i n (\frac{6 x - 4 x}{2}))$

$= (2 s i n \frac{10 x}{2} c o s \frac{2 x}{2}) (2 c o s \frac{10 x}{2} s i n \frac{2 x}{2})$

$= 2 s i n 5 x c o s x \cdot 2 c o s 5 x s i n x$

$= 2 s i n 5 x c o s 5 x \cdot 2 c s i n x c o s x .$

As $s i n 2 θ = 2 s i n θ c o s θ$ we can write.

$L.H.S = s i n (2 * 5 x) = s i n (2 * x)$

$= s i n 10 x s i n 2 x = R.H.S.$ R.H.S

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New answer posted

a year ago

Maths Three Dimensional Geometry

42. If the lines $\frac{x - 3}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} & \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ are perpendicular, find the value of k

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Vishal Baghel

Contributor-Level 10

The direction of ratios of the lines, $\frac{x - 3}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} & \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ , are $- 3, 2 k, 2 a n d 3 k, 1, - 5$ respectively.

It is known that two lines with direction ratios, $a_{1}, b_{1}, c_{1} a n d a_{2}, b_{2}, c_{2}$ , are perpendicular, if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$\begin{array}{l} ∴ - 3 (3 k) + 2 k * 1 + 2 (- 5) = 0 \\ \Rightarrow - 9 k + 2 k - 10 = 0 \\ \Rightarrow 7 k = - 10 \\ \Rightarrow k = \frac{- 10}{7} \end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.

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New answer posted

a year ago

Maths Three Dimensional Geometry

41. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

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Vishal Baghel

Contributor-Level 10

The coordinates of $A, B, C, a n d D a r e (1, 2, 3), (4, 5, 7), ( - 4, 3, - 6),$ and $(2, 9, 2)$ respectively.

The direction ratios of $A B a r e (4 - 1) = 3, (5 - 2) = 3, a n d (7 - 3) = 4$

The direction ratios of $C D a r e (2 - (- 4)) = 6, (9 - 3) = 6, a n d (2 - (- 6)) = 8$

It can be seen that, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = \frac{1}{2}$

Therefore, AB is parallel to CD.

Thus, the angle between $A B a n d C D i s e i t h e r 0 ° o r 18$ $0$ $° .$

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New answer posted

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Maths Three Dimensional Geometry

40. Find the equation of the line parallel to x-axis and passing through the origin.

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Vishal Baghel

Contributor-Level 10

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by $(a, 0, 0), w h e r e a ? R .$ Direction ratios of $O A a r e (a ? 0) = a, 0, 0$

The equation of OA is given by,

$\begin{array}{l} \frac{x ? 0}{a} = \frac{y ? 0}{0} = \frac{z ? 0}{0} \\ ? \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a \end{array}$

Thus, the equation of line parallel to x-axis and passing through origin is

$\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$

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New answer posted

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Maths Trigonometric Functions

$28.$ $c o s (\frac{3 π}{4} + x) - c o s (\frac{3 π}{4} - x) = -\sqrt2 s i n x$

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Payal Gupta

Contributor-Level 10

$28.$ L.H.S = $L.H.S = c o s (\frac{3 π}{4} + x) - c o s (\frac{3 π}{4} - x)$

Using cos (A + B) = cos A cos B – sin A sin B

and cos (A – B) = cos A cos B + sin A sin B

$L.H.S = [c o s \frac{3 π}{4} c o s x - s i n \frac{3 π}{4} s i n x] - [c o s \frac{3 π}{4} c o s x + s i n \frac{3 π}{4} s i n x]$

$= c o s \frac{3 π}{4} c o s x - s i n \frac{3 π}{4} s i n x - c o s \frac{3 π}{4} c o s x - s i n \frac{3 π}{4} s i n x .$

$= - 2 s i n \frac{3 π}{4} s i n x .$

$= - 2 s i n (\frac{4 π - π}{4}) s i n x .$

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New answer posted

a year ago

Maths Three Dimensional Geometry

39. If $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1} .$

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Vishal Baghel

Contributor-Level 10

It is given that $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines. Therefore,

$\begin{array}{l} l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 . . . . . . . . . . (1) \\ {l_{1}}^{2} + {m_{1}}^{2} + n_{1}^{2} = 1 . . . . . . . . . . (2) \\ {l_{2}}^{2} + {m_{2}}^{2} + n_{2}^{2} = 1 . . . . . . . . . . (3) \end{array}$

Let $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines $l_{1}, m_{1}, n_{1} a n d l_{2}, m_{2}, n_{2} .$

$\begin{array}{l} ∴ l l_{1} + m m_{1} + n n_{1} = 0 \\ l l_{2} + m m_{2} + n n_{2} = 0 \\ ∴ \frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} l_{2} - n_{2} l_{1}} = \frac{n}{l_{1} m_{2} - l_{2} m_{1}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{2})}^{2}} \\ = \frac{l^{2} + m^{2} + n^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{2})}^{2}} . . . . . . . . . . (4) \end{array}$

$l, m, n$ are the direction cosines of the line.

$∴ l^{2} + m^{2} + n^{2} = 1 \dots (5)$

It is known that,

$\begin{array}{l} ({l_{1}}^{2} + {m_{1}}^{2} + n_{1}^{2}) ({l_{2}}^{2} + {m_{2}}^{2} + n_{2}^{2}) - {(l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})}^{2} \\ = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} \\ F r o m, (1), (2) & (3), w e . o b t a i n \\ \Rightarrow 1.1 - 0 = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} \\ ∴ {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - l_{2} m_{1})}^{2} = 1 . . . . . . . . . . (6) \end{array}$

Substituting the values from equations (5) and (6) in equation (4), we obtain

$\frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - l_{2} m_{1})}^{2}} = 1$

Thus, the direction cosines of the required line are $m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1}, l_{1} m_{2} - l_{2} m_{1}$

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New answer posted

a year ago

Maths Three Dimensional Geometry

38. Show that the line joining the origin to the point $(2, 1, 1)$ is perpendicular to the line determined by the points $(3, 5, - 1), (4, 3, - 1) .$

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Vishal Baghel

Contributor-Level 10

Let OA be the line joining the origin, $O (0, 0, 0),$ and the point, $A (2, 1, 1) .$

Also, let BC be the line joining the points, $B (3, 5, - 1) a n d C (4, 3, - 1) .$

The direction ratios of $O A a r e 2, 1, a n d 1 a n d o f B C a r e (4 - 3) = 1, (3 - 5) = - 2, a n d (- 1 + 1) = 0$

OA is perpendicular to $B C, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$∴ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 1 + 1 (- 2) + 1 * 0 = 2 - 2 = 0$

Thus, OA is perpendicular to BC.

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New answer posted

a year ago

Maths Three Dimensional Geometry

37. In the following cases find the distances of each of the given points from the corresponding given plane:

$\begin{array}{l} \end{array} (a) P o i n t (0, 0, 0) P l a n e 3 x - 4 y + 12 z = 3$

$(b) P o i n t (3, - 2, 1) P l a n e 2 x - y + 2 z + 3 = 0$

$c) P o i n t (2, 3, - 5) P l a n e x + 2 y - 2 z = 9$

$(d) P o i n t (- 6, 0, 0) P l a n e 2 x - 3 y + 6 z - 2 = 0$

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Vishal Baghel

Contributor-Level 10

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New answer posted

a year ago

Maths Trigonometric Functions

27. sin (n + 1)xsin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

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Payal Gupta

Contributor-Level 10

27. L.H.S $L.HS = s i n (x + 1) x s i n (x + 2) x + c o s (x + 1) x c o s (x + 2) x .$

$= c o s (x + 1) x c o s (x + 2) x + s i n (x + 1) x s i n (x + 2) x .$

Let A = (n+1)x and B = (n+2)x

So, L.H.S = cosAcosB + sin Asin B

$= c o s (A - B)$

Putting values of A and B we get,

L.H.S = $L.H.S = c o s [(n + 1) x - (n + 2) x]$

$= c o s [n x + x - n x - 2 x]$

$= c o s (- x)$

$= c o s x = R.H.S$ R.H.S

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New answer posted

a year ago

Maths Three Dimensional Geometry

36. In the following cases, determine whether the given planes are parallel or perpendicular and in case they are neither, find the angle between them.

$\begin{array}{l} \end{array} (a) 7 x + 5 y + 6 z + 30 = 0 a n d 3 x - y - 10 z + 4 = 0$

$(b) 2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

$(c) 2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

$(d) 2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

$(e) 4 x + 8 y + z - 8 = 0 a n d y + z - 4 = 0$

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Vishal Baghel

Contributor-Level 10

The direction ratios of normal to the plane, $L_{1} : a_{1} x + b_{1} y + c_{1} z = 0 .$ , are $a_{1}, b_{1}, c_{1}$ and $L_{2} : a_{2} x + b_{2} y + c_{2} z = 0 .$

$L_{1}$
$\begin{array}{l} L_{2}, i f \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ L_{1} ⊥ L_{2}, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \end{array}$

The angle between $L_{1} & L_{2}$ is given by,

(b) The equations of the planes are $2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

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

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

Here, $a_{1} = 2, b_{1} = - 2, c_{1} = 4 & a_{2} =, b_{2} = - 3, c_{2} = 6$

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 3 + 1 * (- 2) (- 3) + 4 * 6 = 6 + 6 + 24 = 36 \neq 0$

Thus, the given planes are not perpendicular to each other.

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{- 2}{- 3} = \frac{2}{3} & \frac{c_{1}}{c_{2}} = \frac{4}{6} = \frac{2}{3} \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Thus, the given planes are parallel to each other

(d) The equations of the planes are and $2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

$\begin{array}{l} a_{1} = 2, b_{1} = - 1, c_{1} = 3 & a_{2} =, b_{2} = - 1, c_{2} = 3 \\ \frac{a_{1}}{a_{2}} = \frac{2}{2} = 1, \frac{b_{1}}{b_{2}} = \frac{- 1}{- 1} = 1 & \frac{c_{1}}{c_{2}} = \frac{3}{3} = 1 \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Thus, the given lines are parallel to each other

(e) The equations of the given planes are $4 x + 8 y + z - 8 = 0 a n d y + z - 4 = 0$ $\begin{array}{l} a_{1} = 4, b_{1} = 8, c_{1} = 1 & a_{2} =, b_{2} = 1, c_{2} = 1 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 4 * 0 + 8 * 1 + 1 = 9 \neq 0 \end{array}$

Therefore, the given lines are not perpendicular to each

...more

The direction ratios of normal to the plane, $L_{1} : a_{1} x + b_{1} y + c_{1} z = 0 .$ , are $a_{1}, b_{1}, c_{1}$ and $L_{2} : a_{2} x + b_{2} y + c_{2} z = 0 .$

$L_{1}$
$\begin{array}{l} L_{2}, i f \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ L_{1} ⊥ L_{2}, i f a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \end{array}$

The angle between $L_{1} & L_{2}$ is given by,

(b) The equations of the planes are $2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0$

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

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are $2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1$

Here, $a_{1} = 2, b_{1} = - 2, c_{1} = 4 & a_{2} =, b_{2} = - 3, c_{2} = 6$

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 * 3 + 1 * (- 2) (- 3) + 4 * 6 = 6 + 6 + 24 = 36 \neq 0$

Thus, the given planes are not perpendicular to each other.

Thus, the given planes are parallel to each other

(d) The equations of the planes are and $2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0$

Thus, the given lines are parallel to each other

Therefore, the given lines are not perpendicular to each other.

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{4}{0}, \frac{b_{1}}{b_{2}} = \frac{8}{1} = 8 & \frac{c_{1}}{c_{2}} = \frac{1}{1} = 1 \\ ∴ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \end{array}$

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

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